Freecomplexnumbers

Complex Numbers and the Complex Exponential
1. Complex numbers
The equation x2
+ 1 = 0 has no solutions, because for any real number x the square
x2
is nonnegative, and so x2
+ 1 can never be less than 1. In spite of this it turns out to
be very useful to assume that there is a number i for which one has
(1) i2
= −1.
Any complex number is then an expression of the form a + bi, where a and b are old-
fashioned real numbers. The number a is called the real part of a + bi, and b is called its
imaginary part.
Traditionally the letters z and w are used to stand for complex numbers.
Since any complex number is speciﬁed by two real numbers one can visualize them
by plotting a point with coordinates (a, b) in the plane for a complex number a + bi. The
plane in which one plot these complex numbers is called the Complex plane, or Argand
plane.
z = a + bi
a = Re(z)
b = Im(z)
θ = arg zr = |z| =
√ a
2 + b2
Figure 1. A complex number.
You can add, multiply and divide complex numbers. Here’s how:
To add (subtract) z = a + bi and w = c + di
z + w = (a + bi) + (c + di) = (a + c) + (b + d)i,
z − w = (a + bi) − (c + di) = (a − c) + (b − d)i.
1

2
To multiply z and w proceed as follows:
zw = (a + bi)(c + di)
= a(c + di) + bi(c + di)
= ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
where we have use the defining property i2
= −1 to get rid of i2
.
To divide two complex numbers one always uses the following trick.
a + bi
c + di
=
a + bi
c + di
·
c − di
c − di
=
(a + bi)(c − di)
(c + di)(c − di)
Now
(c + di)(c − di) = c2
− (di)2
= c2
− d2
i2
= c2
+ d2
,
so
a + bi
c + di
=
(ac + bd) + (bc − ad)i
c2 + d2
=
ac + bd
c2 + d2
+
bc − ad
c2 + d2
i
Obviously you do not want to memorize this formula: instead you remember the trick,
i.e. to divide c + di into a + bi you multiply numerator and denominator with c − di.
For any complex number w = c+di the number c−di is called its complex conjugate.
Notation:
w = c + di, ¯w = c − di.
A frequently used property of the complex conjugate is the following formula
(2) w ¯w = (c + di)(c − di) = c2
− (di)2
= c2
+ d2
.
The following notation is used for the real and imaginary parts of a complex number
z. If z = a + bi then
a = the Real Part of z = Re(z), b = the Imaginary Part of z = Im(z).
Note that both Rez and Imz are real numbers. A common mistake is to say that Imz = bi.
The “i” should not be there.
2. Argument and Absolute Value
For any given complex number z = a+bi one defines the absolute value or modulus
to be
|z| = a2 + b2,
so |z| is the distance from the origin to the point z in the complex plane (see figure 1).
The angle θ is called the argument of the complex number z. Notation:
arg z = θ.
The argument is defined in an ambiguous way: it is only defined up to a multiple of
2π. E.g. the argument of −1 could be π, or −π, or 3π, or, etc. In general one says
arg(−1) = π + 2kπ, where k may be any integer.
From trigonometry one sees that for any complex number z = a + bi one has
a = |z| cos θ, and b = |z| sin θ,
so that
|z| = |z| cos θ + i|z| sin θ = |z| cos θ + i sin θ .

3
and
tan θ =
sin θ
cos θ
=
b
a
.
2.1. Example: Find argument and absolute value of z = 2 + i. Solution:
|z| =
√
22 + 12 =
√
5. z lies in the first quadrant so its argument θ is an angle between 0
and π/2. From tan θ = 1
2
we then conclude arg(2 + i) = θ = arctan
1
2
.
3. Geometry of Arithmetic
Since we can picture complex numbers as points in the complex plane, we can also
try to visualize the arithmetic operations “addition” and “multiplication.” To add z and
a
b
c
d
z
w
z + w
Figure 2. Addition of z = a + bi and w = c + di
w one forms the parallelogram with the origin, z and w as vertices. The fourth vertex
then is z + w. See figure 2.
z = a + bi
iz = −b + ai
Figure 3. Multiplication of a + bi by i.
To understand multiplication we first look at multiplication with i. If z = a + bi then
iz = i(a + bi) = ia + bi2
= ai − b = −b + ai.
Thus, to form iz from the complex number z one rotates z counterclockwise by 90 degrees.
See figure 3.
If a is any real number, then multiplication of w = c + di by a gives
aw = ac + adi,

4
−2z
−z
z
2z
3z
Figure 4. Multiplication of a real and a complex number
so aw points in the same direction, but is a times as far away from the origin. If a < 0
then aw points in the opposite direction. See figure 4.
Next, to multiply z = a + bi and w = c + di we write the product as
zw = (a + bi)w = aw + biw.
Figure 5 shows a + bi on the right. On the left, the complex number w was first drawn,
iw
w
aw
aw+biw
a
a+bi
biw
θ
θ
ϕ
b
Figure 5. Multiplication of two complex numbers
then aw was drawn. Subsequently iw and biw were constructed, and finally zw = aw+biw
was drawn by adding aw and biw.
One sees from figure 5 that since iw is perpendicular to w, the line segment from 0
to biw is perpendicular to the segment from 0 to aw. Therefore the larger shaded triangle
on the left is a right triangle. The length of the adjacent side is a|w|, and the length of
the opposite side is b|w|. The ratio of these two lengths is a : b, which is the same as for
the shaded right triangle on the right, so we conclude that these two triangles are similar.
The triangle on the left is |w| times as large as the triangle on the right. The two
angles marked θ are equal.
Since |zw| is the length of the hypothenuse of the shaded triangle on the left, it is |w|
times the hypothenuse of the triangle on the right, i.e. |zw| = |w| · |z|.

5
The argument of zw is the angle θ + ϕ; since θ = arg z and ϕ = arg w we get the
following two formulas
|zw| = |z| · |w|(3)
arg(zw) = arg z + arg w,(4)
in other words,
when you multiply complex numbers, their lengths get multiplied
and their arguments get added.
4. Applications in Trigonometry
4.1. Unit length complex numbers. For any θ the number z = cos θ + i sin θ has
length 1: it lies on the unit circle. Its argument is arg z = θ. Conversely, any complex
number on the unit circle is of the form cos φ + i sin φ, where φ is its argument.
4.2. The Addition Formulas for Sine & Cosine. For any two angles θ and φ one
can multiply z = cos θ+i sin θ and w = cos φ+isin φ. The product zw is a complex number
of absolute value |zw| = |z|·|w| = 1·1, and with argument arg(zw) = arg z+arg w = θ+φ.
So zw lies on the unit circle and must be cos(θ + φ) + i sin(θ + φ). Thus we have
(5) (cos θ + i sin θ)(cos φ + i sin φ) = cos(θ + φ) + i sin(θ + φ).
By multiplying out the Left Hand Side we get
(cos θ + i sin θ)(cos φ + i sin φ) = cos θ cos φ − sin θ sin φ(6)
+ i(sin θ cos φ + cos θ sin φ).
Compare the Right Hand Sides of (5) and (6), and you get the addition formulas for Sine
and Cosine:
cos(θ + φ) = cos θ cos φ − sin θ sin φ
sin(θ + φ) = sin θ cos φ + cos θ sin φ
4.3. De Moivre’s formula. For any complex number z the argument of its square
z2
is arg(z2
) = arg(z · z) = arg z + arg z = 2 arg z. The argument of its cube is arg z3
=
arg(z · z2
) = arg(z) + arg z2
= arg z + 2 arg z = 3 arg z. Continuing like this one ﬁnds that
(7) arg zn
= n arg z
for any integer n.
Applying this to z = cos θ + i sin θ you ﬁnd that zn
is a number with absolute value
|zn
| = |z|n
= 1n
= 1, and argument n arg z = nθ. Hence zn
= cos nθ + i sin nθ. So we
have found
(8) (cos θ + i sin θ)n
= cos nθ + i sin nθ.
This is de Moivre’s formula.
For instance, for n = 2 this tells us that
cos 2θ + i sin 2θ = (cos θ + i sin θ)2
= cos2
θ − sin2
θ + 2i cos θ sin θ.
Comparing real and imaginary parts on left and right hand sides this gives you the double
angle formulas cos θ = cos2
θ − sin2
θ and sin 2θ = 2 sin θ cos θ.
For n = 3 you get, using the Binomial Theorem, or Pascal’s triangle,
(cos θ + i sin θ)3
= cos3
θ + 3i cos2
θ sin θ + 3i2
cos θ sin2
θ + i3
sin3
θ
= cos3
θ − 3 cos θ sin2
θ + i(3 cos2
θ sin θ − sin3
θ)

6
so that
cos 3θ = cos3
θ − 3 cos θ sin2
θ
and
sin 3θ = cos2
θ sin θ − sin3
θ.
In this way it is fairly easy to write down similar formulas for sin 4θ, sin 5θ, etc.. . .
5. Calculus of complex valued functions
A complex valued function on some interval I = (a, b) ⊆ R is a function f : I → C.
Such a function can be written as in terms of its real and imaginary parts,
(9) f(x) = u(x) + iv(x),
in which u, v : I → R are two real valued functions.
One defines limits of complex valued functions in terms of limits of their real and
imaginary parts. Thus we say that
lim
x→x0
f(x) = L
if f(x) = u(x) + iv(x), L = A + iB, and both
lim
x→x0
u(x) = A and lim
x→x1
v(x) = B
hold. From this definition one can prove that the usual limit theorems also apply to
complex valued functions.
5.1. Theorem. If limx→x0 f(x) = L and limx→x0 g(x) = M, then one has
lim
x→x0
f(x) ± g(x) = L ± M,
lim
x→x0
f(x)g(x) = LM,
lim
x→x0
f(x)
g(x)
=
L
M
, provided M = 0.
The derivative of a complex valued function f(x) = u(x)+iv(x) is defined by simply
differentiating its real and imaginary parts:
(10) f (x) = u (x) + iv (x).
Again, one finds that the sum,product and quotient rules also hold for complex valued
functions.
5.2. Theorem. If f, g : I → C are complex valued functions which are differentiable
at some x0 ∈ I, then the functions f±g, fg and f/g are differentiable (assuming g(x0) = 0
in the case of the quotient.) One has
(f ± g) (x0) = f (x0) ± g (x0)
(fg) (x0) = f (x0)g(x0) + f(x0)g (x0)
f
g
(x0) =
f (x0)g(x0) − f(x0)g (x0)
g(x0)2
Note that the chain rule does not appear in this list! See problem 29 for more about the
chain rule.
6. The Complex Exponential Function
We finally give a definition of ea+bi
. First we consider the case a = 0:

7
1
eiθ
= cos θ + i sin θ
θ
Figure 6. Euler’s definition of eiθ
6.1. Definition. For any real number t we set
eit
= cos t + i sin t.
See Figure 6.
6.2. Example. eπi
= cos π + i sin π = −1. This leads to Euler’s famous formula
eπi
+ 1 = 0,
which combines the five most basic quantities in mathematics: e, π, i, 1, and 0.
Reasons why the definition 6.1 seems a good definition.
Reason 1. We haven’t defined eit
before and we can do anything we like.
Reason 2. Substitute it in the Taylor series for ex
:
eit
= 1 + it +
(it)2
2!
+
(it)3
3!
+
(it)4
4!
+ · · ·
= 1 + it −
t2
2!
− i
t3
3!
+
t4
4!
+ i
t5
5!
− · · ·
= 1 − t2
/2! + t4
/4! − · · ·
+ i t − t3
/3! + t5
/5! − · · ·
= cos t + i sin t.
This is not a proof, because before we had only proved the convergence of the Taylor series
for ex
if x was a real number, and here we have pretended that the series is also good if
you substitute x = it.
Reason 3. As a function of t the definition 6.1 gives us the correct derivative.
Namely, using the chain rule (i.e. pretending it still applies for complex functions) we
would get
deit
dt
= ieit
.
Indeed, this is correct. To see this proceed from our definition 6.1:
deit
dt
=
d cos t + i sin t
dt
=
d cos t
dt
+ i
d sin t
dt
= − sin t + i cos t
= i(cos t + i sin t)

8
Reason 4. The formula ex
· ey
= ex+y
still holds. Rather, we have eit+is
= eit
eis
.
To check this replace the exponentials by their definition:
eit
eis
= (cos t + i sin t)(cos s + i sin s) = cos(t + s) + i sin(t + s) = ei(t+s)
.
Requiring ex
· ey
= ex+y
to be true for all complex numbers helps us decide what
ea+bi
shoud be for arbitrary complex numbers a + bi.
6.3. Definition. For any complex number a + bi we set
ea+bi
= ea
· eib
= ea
(cos b + i sin b).
One verifies as above in “reason 3” that this gives us the right behaviour under differen-
tiation. Thus, for any complex number r = a + bi the function
y(t) = ert
= eat
(cos bt + i sin bt)
satisfies
y (t) =
dert
dt
= rert
.
7. Complex solutions of polynomial equations
7.1. Quadratic equations. The well-known quadratic formula tells you that the
equation
(11) ax2
+ bx + c = 0
has two solutions, given by
(12) x± =
−b ±
√
D
2a
, D = b2
− 4ac.
If the coefficients a, b, c are real numbers and if the discriminant D is positive, then this
formula does indeed give two real solutions x+ and x−. However, if D < 0, then there are
no real solutions, but there are two complex solutions, namely
x± =
−b
2a
± i
√
−D
2a
7.2. Example: solve x2
+ 2x + 5 = 0. Solution: Use the quadratic formula, or
complete the square:
x2
+ 2x + 5 = 0
⇐⇒ x2
+ 2x + 1 = −4
⇐⇒ (x + 1)2
= −4
⇐⇒ x + 1 = ±2i
⇐⇒ x = −1 ± 2i.
So, if you allow complex solutions then every quadratic equation has two solutions,
unless the two solutions coincide (the case D = 0, in which there is only one solution.)

9
1
1
2
+ i
2
√
3−1
2
+ i
2
√
3
−1
−1
2
− i
2
√
3 1
2
− i
2
√
3
Figure 7. The sixth roots of 1. There are six of them, and they re arranged in a
regular hexagon.
7.3. Complex roots of a number. For any given complex number w there is a
method of finding all complex solutions of the equation
(13) zn
= w
if n = 2, 3, 4, · · · is a given integer.
To find these solutions you write w in polar form, i.e. you find r > 0 and θ such that
w = reiθ
. Then
z = r1/n
eiθ/n
is a solution to (13). But it isn’t the only solution, because the angle θ for which w = riθ
isn’t unique – it is only determined up to a multiple of 2π. Thus if we have found one
angle θ for which w = riθ
, then we can also write
w = rei(θ+2kπ)
, k = 0, ±1, ±2, · · ·
The nth
roots of w are then
zk = r1/n
ei θ
n
+2 k
n
π
Here k can be any integer, so it looks as if there are infinitely many solutions. However,
if you increase k by n, then the exponent above increases by 2πi, and hence zk does not
change. In a formula:
zn = z0, zn+1 = z1, zn+2 = z2, . . . zk+n = zk
So if you take k = 0, 1, 2, · · · , n − 1 then you have had all the solutions.
The solutions zk always form a regular polygon with n sides.

10
7.4. Example: find all sixth roots of w = 1. We are to solve z6
= 1. First write
1 in polar form,
1 = 1 · e0i
= 1 · e2kπi
, (k = 0, ±1, ±2, . . .).
Then we take the 6th
root and find
zk = 11/6
e2kπi/6
= ekπi/3
, (k = 0, ±1, ±2, . . .).
The six roots are
z0 = 1 z1 = eπi/3
= 1
2
+ i
2
√
3 z2 = e2πi/3
= −1
2
+ i
2
√
3
z3 = −1 z4 = eπi/3
= −1
2
− i
2
√
3 z5 = eπi/3
= 1
2
− i
2
√
3
8. Other handy things you can do with complex numbers
8.1. Partial fractions. Consider the partial fraction decomposition
x2
+ 3x − 4
(x − 2)(x2 + 4)
=
A
x − 2
+
Bx + C
x2 + 4
The coefficient A is easy to find: multiply with x − 2 and set x = 2 (or rather, take the
limit x → 2) to get
A =
22
+ 3 · 2 − 4
22 + 4
= · · · .
Before we had no similar way of finding B and C quickly, but now we can apply the same
trick: multiply with x2
+ 4,
x2
+ 3x − 4
(x − 2)
= Bx + C + (x2
+ 4)
A
x − 2
,
and substitute x = 2i. This make x2
+ 4 = 0, with result
(2i)2
+ 3 · 2i − 4
(2i − 2)
= 2iB + C.
Simplify the complex number on the left:
(2i)2
+ 3 · 2i − 4
(2i − 2)
=
−4 + 6i − 4
−2 + 2i
=
−8 + 6i
−2 + 2i
=
(−8 + 6i)(−2 − 2i)
(−2)2 + 22
=
28 + 4i
8
=
7
2
+
i
2
So we get 2iB + C = 7
2
+ i
2
; since B and C are real numbers this implies
B =
1
4
, C =
7
2
.

11
8.2. Certain trigonometric and exponential integrals. You can compute
I = e3x
cos 2xdx
by integrating by parts twice. You can also use that cos 2x is the real part of e2ix
. Instead
of computing the real integral I, we look at the following related complex integral
J = e3x
e2ix
dx
which we get from I by replacing cos 2x with e2ix
. Since e2ix
= cos 2x + i sin 2x we have
J = e3x
(cos 2x + i sin 2x)dx = e3x
cos 2xdx + i e3x
sin 2xdx
i.e.,
J = I + something imaginary.
The point of all this is that J is easier to compute than I:
J = e3x
e2ix
dx = e3x+2ix
dx = e(3+2i)x
dx =
e(3+2i)x
3 + 2i
+ C
where we have used that
eax
dx =
1
a
eax
+ C
holds even if a is complex is a complex number such as a = 3 + 2i.
To ﬁnd I you have to compute the real part of J, which you do as follows:
e(3+2i)x
3 + 2i
= e3x cos 2x + i sin 2x
3 + 2i
= e3x (cos 2x + i sin 2x)(3 − 2i)
(3 + 2i)(3 − 2i)
= e3x 3 cos 2x + 2 sin 2x + i(· · · )
13
so
e3x
cos 2xdx = e3x 3
13
cos 2x + 2
13
sin 2x + C.
8.3. Complex amplitudes. A harmonic oscillation is given by
y(t) = A cos(ωt − φ),
where A is the amplitude, ω is the frequency, and φ is the phase of the oscillation.
If you add two harmonic oscillations with the same frequency ω, then you get another
harmonic oscillation with frequency ω. You can prove this using the addition formulas for
cosines, but there’s another way using complex exponentials. It goes like this.
Let y(t) = A cos(ωt − φ) and z(t) = B cos(ωt − θ) be the two harmonic oscillations
we wish to add. They are the real parts of
Y (t) = A {cos(ωt − φ) + i sin(ωt − φ)} = Aeiωt−iφ
= Ae−iφ
eiωt
Z(t) = B {cos(ωt − θ) + i sin(ωt − θ)} = Beiωt−iθ
= Be−iθ
eiωt
Therefore y(t) + z(t) is the real part of Y (t) + Z(t), i.e.
y(t) + z(t) = Re Y (t) + Re Z(t) = Re Y (t) + Z(t) .
The quantity Y (t) + Z(t) is easy to compute:
Y (t) + Z(t) = Ae−iφ
eiωt
+ Be−iθ
eiωt
= Ae−iφ
+ Be−iθ
eiωt
.
If you now do the complex addition
Ae−iφ
+ Be−iθ
= Ce−iψ
,

12
w
i.e. you add the numbers on the right, and compute the absolute value C and argument
−ψ of the sum, then we see that Y (t) + Z(t) = Cei(ωt−ψ)
. Since we were looking for the
real part of Y (t) + Z(t), we get
y(t) + z(t) = A cos(ωt − φ) + B cos(ωt − θ) = C cos(ωt − ψ).
The complex numbers Ae−iφ
, Be−iθ
and Ce−iψ
are called the complex amplitudes for the
harmonic oscillations y(t), z(t) and y(t) + z(t).
The recipe for adding harmonic oscillations can therefore be summarized as follows:
Add the complex amplitudes.
9. PROBLEMS
Computing and Drawing Complex Numbers.
1. Compute the following complex numbers
by hand.
Draw all numbers in the complex (or
“Argand”) plane (use graph paper or quad
paper if necessary).
Compute absolute value and argument
of all numbers involved.
i2; i3; i4; 1/i;
(1 + 2i)(2 − i);
(1 + i)(1 + 2i)(1 + 3i);
(1
2
√
2 + i
2
√
2)2; (1
2
+ i
2
√
3)3;
1
1 + i
; 5/(2 − i);
2. [Deriving the addition formula for
tan(θ + φ)] Let θ, φ ∈ − π
2
, π
2
be two an-
gles.
(a) What are the arguments of
z = 1 + i tan θ and w = 1 + i tan φ?
(Draw both z and w.)
(b) Compute zw.

13
(c) What is the argument of zw?
(d) Compute tan(arg zw).
3. Find formulas for cos 4θ, sin 4θ, cos 5θ
and sin 6θ in terms of cos θ and sin θ, by us-
ing de Moivre’s formula.
4. In the following picture draw 2w, 3
4
w, iw,
−2iw, (2 + i)w and (2 − i)w. (Try to make
a nice drawing, use a ruler.)
Make a new copy of the picture, and
draw ¯w, − ¯w and −w.
Make yet another copy of the drawing.
Draw 1/w, 1/ ¯w, and −1/w. For this draw-
ing you need to know where the unit circle is
in your drawing: Draw a circle centered at
the origin with radius of your choice, and let
this be the unit circle. [Depending on which
circle you draw you will get a different an-
swer!]
5. Verify directly from the definition of addi-
tion and multiplication of complex numbers
that
(a) z + w = w + z
(b) zw = wz
(c) z(v + w) = zv + zw
holds for all complex numbers v, w, and z.
6. True or False? (In mathematics this
means that you should either give a proof
that the statement is always true, or else
give a counterexample, thereby showing that
the statement is not always true.)
For any complex numbers z and w one
has
(a) Re(z) + Re(w) = Re(z + w)
(b) z + w = ¯z + ¯w
(c) Im(z) + Im(w) = Im(z + w)
(d) zw = (¯z)( ¯w)
(e) Re(z)Re(w) = Re(zw)
(f) z/w = (¯z)/( ¯w)
(g) Re(iz) = Im(z)
(h) Re(iz) = iRe(z)
(i) Re(iz) = Im(z)
(j) Re(iz) = iIm(z)
(k) Im(iz) = Re(z)
(l) Re(¯z) = Re(z)
7. The imaginary part of a complex num-
ber is known to be twice its real part. The
absolute value of this number is 4. Which
number is this?
8. The real part of a complex number is
known to be half the absolute value of that
number. The imaginary part of the number
is 1. Which number is it?
The Complex Exponential.
9. Compute and draw the following num-
bers in the complex plane
eπi/3; eπi/2;
√
2e3πi/4; e17πi/4.
eπi + 1; ei ln 2.
1
eπi/4
;
e−πi
eπi/4
;
e2−πi/2
eπi/4
e2009πi; e2009πi/2 .
−8e4πi/3 ; 12eπi + 3e−πi.
10. Compute the absolute value and argu-
ment of e(ln 2)(1+i).
11. Suppose z can be any complex number.
(a) Is it true that ez is always a positive
number?
(b) Is it true that ez = 0?
12. Verify directly from the definition that
e−it
=
1
eit
holds for all real values of t.
13. Show that
cos t =
eit + e−it
2
, sin t =
eit − e−it
2i
14. Show that
cosh x = cos ix, sinh x =
1
i
sin ix.
15. The general solution of a second order
linear differential equation contains expres-
sions of the form Aeiβt +Be−iβt. These can
be rewritten as C1 cos βt + C2 sin βt.
If Aeiβt + Be−iβt = 2 cos βt + 3 sin βt,
then what are A and B?
16. (a) Show that you can write a “cosine-
wave” with amplitude A and phase φ as fol-
lows
A cos(t − φ) = Re zeit
,
where the “complex amplitude” is given by
z = Ae−iφ. (See §8.3).

14
(b) Show that a “sine-wave” with am-
plitude A and phase φ as follows
A sin(t − φ) = Re zeit
,
where the “complex amplitude” is given by
z = −iAe−iφ.
17. Find A and φ where A cos(t − φ) =
2 cos(t) + 2 cos(t − 2
3
π).
12 cos(t − 1
6
π) + 12 sin(t − 1
3
π).
12 cos(t − π/6) + 12 cos(t − π/3).
20. Find A and φ such that A cos(t − φ) =
cos t − 1
6
π +
√
3 cos t − 2
3
π .
Real and Complex Solutions of Algebraic Equations.
21. Find and draw all real and complex so-
lutions of
(a) z2 + 6z + 10 = 0
(b) z3 + 8 = 0
(c) z3 − 125 = 0
(d) 2z2 + 4z + 4 = 0
(e) z4 + 2z2 − 3 = 0
(f) 3z6 = z3 + 2
(g) z5 − 32 = 0
(h) z5 − 16z = 0
Calculus of Complex Valued Functions.
22. Compute the derivatives of the following
functions
f(x) =
1
x + i
g(x) = log x + i arctan x
h(x) = eix2
k(x) = log
i + x
i − x
Try to simplify your answers.
23. (a) Compute
cos 2x
4
dx
by using cos θ = 1
2
(eiθ + e−iθ) and expand-
ing the fourth power.
(b) Assuming a ∈ R, compute
e−2x
sin ax
2
dx.
(same trick: write sin ax in terms of complex
exponentials; make sure your ﬁnal answer
has no complex numbers.)
24. Use cos α = (eiα + e−iα)/2, etc. to eval-
uate these indeﬁnite integrals:
(a) cos2
x dx
(b) cos4
x dx,
(c) cos2
x sin x dx,
(d) sin3
x dx,
(e) cos2
x sin2
x dx,
(f) sin6
x dx
(g) sin(3x) cos(5x) dx
(h) sin2
(2x) cos(3x) dx
(i)
π/4
0
sin(3x) cos(x) dx
(j)
π/3
0
sin3
(x) cos2
(x) dx
(k)
π/2
0
sin2
(x) cos2
(x) dx
(l)
π/3
0
sin(x) cos2
(x) dx
25. Compute the following integrals when
m = n are distinct integers.
(a)
2π
0
sin(mx) cos(nx) dx
(b)
2π
0
sin(nx) cos(nx) dx
(c)
2π
0
cos(mx) cos(nx) dx
(d)
π
0
cos(mx) cos(nx) dx
(e)
2π
0
sin(mx) sin(nx) dx
(f)
π
0
sin(mx) sin(nx) dx
These integrals are basic to the the-
ory of Fourier series, which occurs in

15
many applications, especially in the study of
wave motion (light, sound, economic cycles,
clocks, oceans, etc.). They say that different
frequency waves are “independent”.
26. Show that cos x+sin x = C cos(x+β) for
suitable constants C and β and use this to
evaluate the following integrals.
(a)
dx
cos x + sin x
(b)
dx
(cos x + sin x)2
(c)
dx
A cos x + B sin x
where A and B are any constants.
27. Compute the integrals
π/2
0
sin2
kx sin2
lx dx,
where k and l are positive integers.
28. Show that for any integers k, l, m
π
0
sin kx sin lx sin mx dx = 0
if and only if k + l + m is even.
29. (i) Prove the following version of the
Chain rule: If f : I → C is a differentiable
complex valued function, and g : J → I
is a differentiable real valued function, then
h = f ◦g : J → C is a differentiable function,
and one has
h (x) = f (g(x))g (x).
(ii) Let n ≥ 0 be a nonnegative integer.
Prove that if f : I → C is a differentiable
function, then g(x) = f(x)n is also differen-
tiable, and one has
g (x) = nf(x)n−1
f (x).
Note that the chain rule from part (a) does
not apply! Why?

16
Answers and Hints
(2) (a) arg(1 + i tan θ) = θ + 2kπ, with k any integer.
(b) zw = 1 − tan θ tan φ + i(tan θ + tan φ)
(c) arg(zw) = arg z + arg w = θ + φ (+ a multiple of 2π.)
(d) tan(arg zw) = tan(θ + φ) on one hand, and tan(arg zw) =
tan θ + tan φ
1 − tan θ tan φ
on the other hand.
The conclusion is that
tan(θ + φ) =
tan θ + tan φ
1 − tan θ tan φ
(3) cos 4θ = real part of (cos θ + i sin θ)4. Expand, using Pascal’s triangle to get
cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ.
sin 4θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ.
cos 5θ = cos5 θ − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ
sin 6θ = 6 cos5 θ sin θ − 20 cos3 θ sin3
θ + 6 cos θ sin5
θ.
(6) To prove or disprove the statements set z = a + bi, w = c + di and substitute in the equation.
Then compare left and right hand sides.
(a) Re(z) + Re(w) = Re(z + w) TRUE, because:
Re(z + w) = Re(a + bi + c + di) = Re[(a + c) + (b + d)i] = a + c and
Re(z) + Re(w) = Re(a + bi) + Re(c + di) = a + c.
The other proofs go along the same lines.
(b) z + w = ¯z + ¯w TRUE. Proof: if z = a + bi and w = c + di with a, b, c, d real numbers, then
Re(z) = a, Re(w) = c =⇒ Re(z) + Re(w) = a + c
z + w = a + c + (b + d)i =⇒ Re(z + w) = a + c.
So you see that Re(z) + Re(w) and Re(z + w) are equal.
(c) Im(z) + Im(w) = Im(z + w) TRUE. Proof: if z = a + bi and w = c + di with a, b, c, d real
numbers, then
Im(z) = b, Im(w) = d =⇒ Im(z) + Im(w) = b + d
z + w = a + c + (b + d)i =⇒ Im(z + w) = b + d.
So you see that Im(z) + Im(w) and Im(z + w) are equal.
(d) zw = (¯z)( ¯w) TRUE
(e) Re(z)Re(w) = Re(zw) FALSE. Counterexample: Let z = i and w = i. Then Re(z)Re(w) =
0 · 0 = 0, but Re(zw) = Re(i · i) = Re(−1) = −1.
(f) z/w = (¯z)/( ¯w) TRUE
(g) Re(iz) = Im(z) FALSE (almost true though, only oﬀ by a minus sign)
(h) Re(iz) = iRe(z) FALSE. The left hand side is a real number, the right hand side is an
imaginary number: they can never be equal (except when z = 0.)
(i) Re(iz) = Im(z) same as (g), sorry.
(j) Re(iz) = iIm(z) FALSE
(k) Im(iz) = Re(z) TRUE
(l) Re(¯z) = Re(z) TRUE

17
(7) The number is either 1
5
√
5 + 2
5
i
√
5 or − 1
5
√
5 − 2
5
i
√
5.
(8) ’t is 1
3
√
3 + i.
(10) e(ln 2)(1+i) = eln 2+i ln 2 = eln 2(cos ln 2 + i sin ln 2) so the real part is 2 cos ln 2 and the imag-
inary part is 2 sin ln 2.
(11) ez can be negative, or any other complex number except zero.
If z = x + iy then ez = ex(cos y + i sin y), so the absolute value and argument of ez are |z| = ex
and arg ez = y. Therefore the argument can be anything, and the absolute value can be any
positive real number, but not 0.
(12)
1
eit
=
1
cos t + i sin t
=
1
cos t + i sin t
cos t − i sin t
cos t − i sin t
=
cos t − i sin t
cos2 t + sin2 t
= cos t − i sin t = e−it
.
(15) Aeiβt + Be−iβt = A(cos βt + i sin βt) + B(cos βt − i sin βt) = (A + B) cos βt + i(A − B) sin βt.
So Aeiβt + Be−iβt = 2 cos βt + 3 sin βt holds if A + B = 2, i(A − B) = 3. Solving these two
equations for A and B we get A = 1 − 3
2
i, B = 1 + 3
2
i.
(21) (a) z2 + 6z + 10 = (z + 3)2 + 1 = 0 has solutions z = −3 ± i.
(b) z3 + 8 = 0 ⇐⇒ z3 = −8. Since −8 = 8eπi+2kπ we find that z = 81/3e
π
3
i+ 2
3
kπi
(k any
integer). Setting k = 0, 1, 2 gives you all solutions, namely
k = 0 : z = 2e
π
3
i
= 1 + i
√
3
k = 1 : z = 2e
π
3
i+2πi/3
= −2
k = 2 : z = 2e
π
3
i+4πi/3
= 1 − i
√
3
(c) z3 − 125 = 0: z0 = 5, z1 = − 5
2
+ 5
2
i
√
3, z2 = − 5
2
− 5
2
i
√
3
(d) 2z2 + 4z + 4 = 0: z = −1 ± i.
(e) z4 + 2z2 − 3 = 0: z2 = 1 or z2 = −3, so the four solutions are ±1, ±i
√
3.
(f) 3z6 = z3 + 2: z3 = 1 or z3 = − 2
3
. The six solutions are therefore
− 1
2
± i
2
√
3, 1 (from z3
= 1)
− 3 2
3
, 3 2
3
1
2
± i
2
√
3 , (from z3
= − 2
3
)
(g) z5 − 32 = 0: The five solutions are
2, 2 cos 2
5
π ± 2i sin 2
5
π, 2 cos 4
5
π ± 2i sin 4
5
π.
Note that 2 cos 6
5
π+2i sin 6
5
π = 2 cos 4
5
π−2i sin 4
5
π, and likewise, 2 cos 8
5
π+2i sin 8
5
π = 2 cos 2
5
π−
2i sin 2
5
π. (Make a drawing of these numbers to see why).
(h) z5 − 16z = 0: Clearly z = 0 is a solution. Factor out z to find the equation z4 − 16 = 0 whose
solutions are ±2, ±2i. So the five solutions are 0, ±2, and ±2i
(22) f (x) = −1
(x+i)2 . In this computation you use the quotient rule, which is valid for complex
valued functions.
g (x) = 1
x
+ i
1+x2
h (x) = 2ixeix2
. Here we are allowed to use the Chain Rule because h(x) is of the form h1(h2(x)),
where h1(y) = eiy is a complex valued function of a real variable, and h2(x) = x2 is a real valued
function of a real variable (a “221 function”).
(23) (a) Use the hint:
cos 2x
4
dx =
e2ix + e−2ix
2
4
dx
= 1
16
e2ix
+ e−2ix 4
dx

18
The fourth line of Pascal’s triangle says (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4. Apply this with
a = e2ix, b = e−2ix and you get
cos 2x
4
dx =
1
16
e8ix
+ 4e4ix
+ 6 + 4e−4ix
+ e−8ix
dx
=
1
16
1
8i
e8ix
+
4
4i
e4ix
+ 6x +
4
−4i
e−4ix
+
1
−8i
e−8ix
+ C.
We could leave this as the answer since we’re done with the integral. However, we are asked to
simplify our answer, and since we know ahead of time that the answer is a real function we should
rewrite this as a real function. There are several ways of doing this, one of which is to carefully
match complex exponential terms with their complex conjugates (e.g. e8ix with e−8ix.) This gives
us
cos 2x
4
dx =
1
16
e8ix − e−8ix
8i
+
e4ix − e−4ix
i
+ 6x + C.
Finally, we use the formula sin θ = eiθ
−e−iθ
2i
to remove the complex exponentials. We end up
with the answer
cos 2x
4
dx =
1
16
1
4
sin 8x + 2 sin 4x + 6x + C = 1
64
sin 8x + 1
8
sin 4x + 3
8
x + C.
(b) Use sin θ = (eiθ − e−iθ)/(2i):
e−2x
sin ax
2
dx = e−2x eiax − e−iax
2i
2
dx
=
1
(2i)2
e−2x
e2iax
− 2 + e−2iax
dx
= − 1
4
e(−2+2ia)x
− 2 + e(−2−2ia)x
dx
= − 1
4
e(−2+2ia)x
−2 + 2ia
A
−2x +
e(−2−2ia)x
−2 − 2ia
B
+ C.(†)
We are done with integrating. The answer must be a real function (being the integral of a real
function), so we have to be able to write our answer in a real form. To get this real form we must
expand the complex exponentials above, and do the division by −2 + 2ia and −2 − 2ia. This is
still a fair amount of work, but we can cut the amount of work in half by noting that the terms
A and B are complex conjugates of each other, i.e. they are the same, except for the sign in front
of i: you get B from A by changing all i’s to −i’s. So once we have simpliﬁed A we immediately
know B.
We compute A as follows
A =
−2 − 2ia
(−2 − 2ia)(−2 + 2ia)
e−2x+2iax
=
(−2 − 2ia)e−2x(cos 2ax + i sin 2ax)
(−2)2 + (−2a)2
=
e−2x
4 + 4a2
(−2 cos 2ax + 2a sin 2ax) + i
e−2x
4 + 4a2
(−2a cos 2ax − 2 sin 2ax).
Hence
B =
e−2x
4 + 4a2
(−2 cos 2ax + 2a sin 2ax) − i
e−2x
4 + 4a2
(−2a cos 2ax − 2 sin 2ax).
and
A + B =
2e−2x
4 + 4a2
(−2 cos 2ax + 2a sin 2ax) =
e−2x
1 + a2
(− cos 2ax + a sin 2ax).
Substitute this in (†) and you get the real form of the integral
e−2x
sin ax
2
dx = − 1
4
e−2x
1 + a2
(− cos 2ax + a sin 2ax) +
x
2
+ C.

19
(24) (a) This one can be done with the double angle formula, but if you had forgotten that,
complex exponentials work just as well:
cos2
x dx =
eix + e−ix
2
2
dx
= 1
4
e2ix
+ 2 + e−2ix
dx
= 1
4
1
2i
e2ix
+ 2x + 1
−2i
e−2ix
+ C
= 1
4
e2ix − e−2ix
2i
+ 2x + C
= 1
4
sin 2x + 2x + C
= 1
4
sin 2x +
x
2
+ C.
(c), (d) using complex exponentials works, but for these integrals substituting u = sin x works
better, if you use cos2 x = 1 − sin2 x.
(e) Use (a − b)(a + b) = a2 − b2 to compute
cos2
x sin2
x =
(eix + e−ix)2
22
(eix − e−ix)2
(2i)2
= 1
−16
(e2ix
+ e−2ix
)2
= 1
−16
(e4ix
+ 2 + e−4ix
)
First variation: The integral is
cos2
x sin2
x dx = 1
−16
( 1
4i
e4ix
+ 2x + 1
−4i
e−4ix
) + C = 1
−32
sin 4x − 1
8
x + C.
Second variation: Get rid of the complex exponentials before integrating:
1
−16
(e4ix
+ 2 + e−4ix
) = 1
−16
(2 cos 4x + 2) = − 1
8
(cos 4x + 1),
If you integrate this you get the same answer as above.
(j) and (l): Substituting complex exponentials will get you the answer, but for these two
integrals you’re much better oﬀ substituting u = cos x (and keep in mind that sin2 x = 1−cos2 x.)
(k) See (e) above.

20
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location until at least one year after the last time you
distribute an Opaque copy (directly or through your
agents or retailers) of that edition to the public.
It is requested, but not required, that you contact the
authors of the Document well before redistributing any
large number of copies, to give them a chance to pro-
vide you with an updated version of the Document.
4. MODIFICATIONS
You may copy and distribute a Modified Version of
the Document under the conditions of sections 2 and 3
above, provided that you release the Modified Version
under precisely this License, with the Modified Ver-
sion filling the role of the Document, thus licensing
distribution and modification of the Modified Version
to whoever possesses a copy of it. In addition, you
must do these things in the Modified Version:
A. Use in the Title Page (and on the cov-
ers, if any) a title distinct from that of
the Document, and from those of previ-
ous versions (which should, if there were
any, be listed in the History section of the
Document). You may use the same title
as a previous version if the original pub-
lisher of that version gives permission.
B. List on the Title Page, as authors, one
or more persons or entities responsible
for authorship of the modifications in the
Modified Version, together with at least
five of the principal authors of the Doc-
ument (all of its principal authors, if it
has fewer than five), unless they release
you from this requirement.
C. State on the Title page the name of the
publisher of the Modified Version, as the
publisher.
D. Preserve all the copyright notices of the
Document.
E. Add an appropriate copyright notice for
your modifications adjacent to the other
copyright notices.
F. Include, immediately after the copyright
notices, a license notice giving the pub-
lic permission to use the Modified Ver-
sion under the terms of this License, in
the form shown in the Addendum below.
G. Preserve in that license notice the full
lists of Invariant Sections and required
Cover Texts given in the Document’s li-
cense notice.
H. Include an unaltered copy of this License.
I. Preserve the section Entitled “History”,
Preserve its Title, and add to it an item
stating at least the title, year, new au-
thors, and publisher of the Modified Ver-
sion as given on the Title Page. If there is
no section Entitled “History” in the Doc-
ument, create one stating the title, year,
authors, and publisher of the Document
as given on its Title Page, then add an
item describing the Modified Version as
stated in the previous sentence.
J. Preserve the network location, if any,
given in the Document for public access
to a Transparent copy of the Document,
and likewise the network locations given
in the Document for previous versions it
was based on. These may be placed in
the “History” section. You may omit a
network location for a work that was pub-
lished at least four years before the Docu-
ment itself, or if the original publisher of
the version it refers to gives permission.
K. For any section Entitled “Acknowledge-
ments” or “Dedications”, Preserve the
Title of the section, and preserve in
the section all the substance and tone
of each of the contributor acknowledge-
ments and/or dedications given therein.
L. Preserve all the Invariant Sections of the
Document, unaltered in their text and
in their titles. Section numbers or the
equivalent are not considered part of the
section titles.
M. Delete any section Entitled “Endorse-
ments”. Such a section may not be in-
cluded in the Modified Version.
N. Do not retitle any existing section to be
Entitled “Endorsements” or to conflict in
title with any Invariant Section.
O. Preserve any Warranty Disclaimers.
If the Modified Version includes new front-matter sec-
tions or appendices that qualify as Secondary Sections
and contain no material copied from the Document,
you may at your option designate some or all of these
sections as invariant. To do this, add their titles to
the list of Invariant Sections in the Modified Version’s
license notice. These titles must be distinct from any
other section titles.
You may add a section Entitled “Endorsements”, pro-
vided it contains nothing but endorsements of your
Modified Version by various parties—for example,
statements of peer review or that the text has been
approved by an organization as the authoritative def-
inition of a standard.
You may add a passage of up to five words as a Front-
Cover Text, and a passage of up to 25 words as a
Back-Cover Text, to the end of the list of Cover Texts
in the Modified Version. Only one passage of Front-
Cover Text and one of Back-Cover Text may be added
by (or through arrangements made by) any one entity.
If the Document already includes a cover text for the
same cover, previously added by you or by arrange-
ment made by the same entity you are acting on be-
half of, you may not add another; but you may replace
the old one, on explicit permission from the previous
publisher that added the old one.
The author(s) and publisher(s) of the Document do
not by this License give permission to use their names
for publicity for or to assert or imply endorsement of
any Modified Version.
5. COMBINING DOCUMENTS
You may combine the Document with other documents
released under this License, under the terms defined
in section 4 above for modified versions, provided that
you include in the combination all of the Invariant Sec-
tions of all of the original documents, unmodified, and
list them all as Invariant Sections of your combined
work in its license notice, and that you preserve all
their Warranty Disclaimers.
The combined work need only contain one copy of this
License, and multiple identical Invariant Sections may
be replaced with a single copy. If there are multiple
Invariant Sections with the same name but different
contents, make the title of each such section unique by
adding at the end of it, in parentheses, the name of the
original author or publisher of that section if known,
or else a unique number. Make the same adjustment
to the section titles in the list of Invariant Sections in
the license notice of the combined work.

22
In the combination, you must combine any sections
Entitled “History” in the various original documents,
forming one section Entitled “History”; likewise com-
bine any sections Entitled “Acknowledgements”, and
any sections Entitled “Dedications”. You must delete
all sections Entitled “Endorsements”.
6. COLLECTIONS OF DOCUMENTS
You may make a collection consisting of the Document
and other documents released under this License, and
replace the individual copies of this License in the var-
ious documents with a single copy that is included in
the collection, provided that you follow the rules of
this License for verbatim copying of each of the docu-
ments in all other respects.
You may extract a single document from such a collec-
tion, and distribute it individually under this License,
provided you insert a copy of this License into the ex-
tracted document, and follow this License in all other
respects regarding verbatim copying of that document.
7. AGGREGATION WITH INDEPENDENT WORKS
A compilation of the Document or its derivatives with
other separate and independent documents or works,
in or on a volume of a storage or distribution medium,
is called an “aggregate” if the copyright resulting from
the compilation is not used to limit the legal rights
of the compilation’s users beyond what the individ-
ual works permit. When the Document is included in
an aggregate, this License does not apply to the other
works in the aggregate which are not themselves de-
rivative works of the Document.
If the Cover Text requirement of section 3 is applica-
ble to these copies of the Document, then if the Doc-
ument is less than one half of the entire aggregate,
the Document’s Cover Texts may be placed on cov-
ers that bracket the Document within the aggregate,
or the electronic equivalent of covers if the Document
is in electronic form. Otherwise they must appear on
printed covers that bracket the whole aggregate.
8. TRANSLATION
Translation is considered a kind of modification, so you
may distribute translations of the Document under the
terms of section 4. Replacing Invariant Sections with
translations requires special permission from their
copyright holders, but you may include translations
of some or all Invariant Sections in addition to the
original versions of these Invariant Sections. You may
include a translation of this License, and all the li-
cense notices in the Document, and any Warranty Dis-
claimers, provided that you also include the original
English version of this License and the original ver-
sions of those notices and disclaimers. In case of a
disagreement between the translation and the original
version of this License or a notice or disclaimer, the
original version will prevail.
If a section in the Document is Entitled “Acknowledge-
ments”, “Dedications”, or “History”, the requirement
(section 4) to Preserve its Title (section 1) will typi-
cally require changing the actual title.
9. TERMINATION
You may not copy, modify, sublicense, or distribute
the Document except as expressly provided under this
License. Any attempt otherwise to copy, modify, sub-
license, or distribute it is void, and will automatically
terminate your rights under this License.
However, if you cease all violation of this License, then
your license from a particular copyright holder is rein-
stated (a) provisionally, unless and until the copyright
holder explicitly and finally terminates your license,
and (b) permanently, if the copyright holder fails to
notify you of the violation by some reasonable means
prior to 60 days after the cessation.
Moreover, your license from a particular copyright
holder is reinstated permanently if the copyright
holder notifies you of the violation by some reasonable
means, this is the first time you have received notice
of violation of this License (for any work) from that
copyright holder, and you cure the violation prior to
30 days after your receipt of the notice.
Termination of your rights under this section does not
terminate the licenses of parties who have received
copies or rights from you under this License. If your
rights have been terminated and not permanently re-
instated, receipt of a copy of some or all of the same
material does not give you any rights to use it.
10. FUTURE REVISIONS OF THIS LICENSE
The Free Software Foundation may publish new, re-
vised versions of the GNU Free Documentation License
from time to time. Such new versions will be sim-
ilar in spirit to the present version, but may differ
in detail to address new problems or concerns. See
http://www.gnu.org/copyleft/.
Each version of the License is given a distinguishing
version number. If the Document specifies that a par-
ticular numbered version of this License “or any later
version” applies to it, you have the option of follow-
ing the terms and conditions either of that specified
version or of any later version that has been published
(not as a draft) by the Free Software Foundation. If
the Document does not specify a version number of
this License, you may choose any version ever pub-
lished (not as a draft) by the Free Software Founda-
tion. If the Document specifies that a proxy can de-
cide which future versions of this License can be used,
that proxy’s public statement of acceptance of a ver-
sion permanently authorizes you to choose that version
for the Document.
11. RELICENSING
“Massive Multiauthor Collaboration Site” (or “MMC
Site”) means any World Wide Web server that pub-
lishes copyrightable works and also provides promi-
nent facilities for anybody to edit those works. A pub-
lic wiki that anybody can edit is an example of such
a server. A “Massive Multiauthor Collaboration” (or
“MMC”) contained in the site means any set of copy-
rightable works thus published on the MMC site.
“CC-BY-SA” means the Creative Commons
Attribution-Share Alike 3.0 license published by Cre-
ative Commons Corporation, a not-for-profit corpora-
tion with a principal place of business in San Fran-
cisco, California, as well as future copyleft versions of
that license published by that same organization.
“Incorporate” means to publish or republish a Docu-
ment, in whole or in part, as part of another Docu-
ment.
An MMC is “eligible for relicensing” if it is licensed un-
der this License, and if all works that were first pub-
lished under this License somewhere other than this
MMC, and subsequently incorporated in whole or in
part into the MMC, (1) had no cover texts or invari-
ant sections, and (2) were thus incorporated prior to
November 1, 2008.
The operator of an MMC Site may republish an MMC
contained in the site under CC-BY-SA on the same site
at any time before August 1, 2009, provided the MMC
is eligible for relicensing.

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