MIT Math Syllabus 10-3 Lesson 5: Complex numbers

INTRODUCTION TO COMPLEX NUMBERS

Recall that = 3 because 32 = 9.
Now consider the expression .
To find , we need to find a number c such that c2 = –9.
However, the square of any real number c (except zero) is a
positive number.
Consequently, we must expand our concept of number to
include numbers whose squares are negative numbers.
Around the seventeenth century, a new number, called an
imaginary number, was defined so that a negative number would
have a square root.
9
9
9

The letter i was chosen to represent the number whose square is
–1.
Definition of i
The imaginary unit, designated by the letter i, is the number
such that i 2 = –1.
The principal square root of a negative number is defined in
terms of i.

Definition of an Imaginary Number
If a is a positive real number, then The number
is called an imaginary number.
EXAMPLE
It is customary to write i in front of a radical sign, as we did
for , to avoid confusing with
.aia  ai

Definition of a Complex Number
A complex number is a number of the form a + bi where
a and b are real numbers and The number a is the real
part of a + bi, and b is the imaginary part.
EXAMPLE
–3 + 5i
2 – 6i
5
7i
Real part: –3; imaginary part: 5
Real part: 2; imaginary part: –6
Real part: 5; imaginary part: 0
Real part: 0; imaginary part: 7
.1i

Note from these examples that a real number is a complex
number whose imaginary part is zero, and an imaginary number
is a complex number whose real part is zero and whose
imaginary part is not zero.
Note from the diagram that the
set of real numbers is a subset
of the complex numbers and the
set of imaginary numbers is a
separate subset of the complex
numbers.

The set of real numbers and the set of imaginary numbers are
disjoint sets.
The example in slide #7 illustrates how to write a complex
number in the standard form a + bi.

Write in the form a + bi.
Solution:
WRITE A COMPLEX NUMBER IN STANDARD FORM
EXAMPLE

ADDITION AND SUBTRACTION OF COMPLEX NUMBERS

All the standard arithmetic operations that are applied to real
numbers can be applied to complex numbers.
Definition of Addition and Subtraction of Complex Numbers
If a + bi and c + di are complex numbers, then
Addition (a + bi ) + (c + di ) = (a + c ) + (b + d )i
Subtraction (a + bi ) – (c + di ) = (a – c ) + (b – d )i

Basically, these rules say that to add two complex numbers, add
the real parts and add the imaginary parts.
To subtract two complex numbers, subtract the real parts and
subtract the imaginary parts.

Simplify. a. (7 – 2i ) + (–2 + 4i ) b. (–9 + 4i ) – (2 – 6i )
Solution:
a. (7 – 2i ) + (–2 + 4i ) = (7 + (–2)) + (–2 + 4)i
= 5 + 2i
b. (–9 + 4i ) – (2 – 6i ) = (–9 – 2) + (4 – (–6))i
= –11 + 10i
ADD OR SUBTRACT COMPLEX NUMBERS
EXAMPLE 2

MULTIPLICATION OF COMPLEX NUMBERS

When multiplying complex numbers, the term i 2 is frequently a
part of the product. Recall that i 2 = –1.
Therefore,
3i(5i ) = 15i 2 = 15(–1) = –15
–2i(6i ) = –12i 2 = –12(–1) = 12
4i(3 – 2i ) = 12i – 8i 2 = 12i – 8(–1) = 8 + 12i

When multiplying square roots of negative numbers, first rewrite
the radical expressions using i. For instance,
Note from this example that it would have been incorrect to
multiply the radicands of the two radical expressions. To
illustrate:

To multiply two complex numbers, we use the following
definition.
Definition of Multiplication of Complex Numbers
If a + bi and c + di are complex numbers, then
(a + bi )(c + di ) = (ac – bd ) + (ad + bc )i

Because every complex number can be written as a sum of two
terms, it is natural to perform multiplication on complex
numbers in a manner consistent with the operation defined on
binomials and the definition i 2 = –1.
By using this analogy, you can multiply complex numbers without
memorizing the definition.

Multiply.
a. 3i(2 – 5i )
b. (3 – 4i )(2 + 5i )
Solution:
a. 3i(2 – 5i ) = 6i – 15i 2
= 6i – 15(–1)
= 15 + 6i
Replace i2 with –1.
Write in standard form.

b. (3 – 4i )(2 + 5i ) = 6 + 15i – 8i – 20i 2
= 6 + 15i – 8i – 20(–1)
= 6 + 15i – 8i + 20
= 26 + 7i
Replace i2 with –1.
Simplify.
Write in standard form.
cont’d

DIVISION OF COMPLEX NUMBERS
Recall that the number is not in simplest form because there
is a radical expression in the denominator.
Similarly, is not in simplest form because
To write this expression in simplest form, multiply the numerator
and denominator by i.

Here is another example.
Recall that to simplify the quotient , we multiply the
numerator and denominator by the conjugate of ,
which is
In a similar manner, to find the quotient of two complex
numbers, we multiply the numerator and denominator by the
conjugate of the denominator.

The complex numbers a + bi and a – bi are called complex
conjugates or conjugates of each other.
The conjugate of the complex number z is denoted by
For instance,
and

Consider the product of a complex number and its conjugate. For
instance,
(2 + 5i )(2 – 5i ) = 4 – 10i + 10i – 25i 2
= 4 – 25(–1)
= 4 + 25
= 29
Note that the product is a real number. This is always true.

Product of Complex Conjugates
The product of a complex number and its conjugate is a real
number. That is,
(a + bi )(a – bi ) = a2 + b2.
EXAMPLE
(5 + 3i )(5 – 3i ) = 52 + 32 = 25 + 9 = 34
The next example shows how the quotient of two complex
numbers is determined by using conjugates.

Simplify:
Solution:
Multiply the numerator
and
denominator by the
conjugate
of the denominator.

cont’d

POWERS OF i
The following powers of i illustrate a pattern:
i 1 = i
i 2 = –1
i 3 = i 2  i = (–1)i = –i
i 4 = i 2  i 2 = (–1)(–1) = 1
i 5 = i 4  i = 1  i = i
i 6 = i 4  i 2 = 1(–1) = –1
i 7 = i 4  i 3 = 1(–i ) = –i
i 8 = (i 4)2 = 12 = 1
Because i 4 = 1, (i 4)n = 1n = 1 for any integer n.

POWERS OF i
Thus it is possible to evaluate powers of i by factoring out
powers of i 4, as shown in the following.
i 27 = (i 4)6  i 3 = 16  i 3 = 1  (–i ) = –i
The following theorem can also be used to evaluate powers of i.
Powers of i
If n is a positive integer, then i n = i r, where r is the remainder of
the division of n by 4.

POWERS OF i
Evaluate: i 153
Solution:
Use the powers of i theorem.
i 153 = i 1
= i
Remainder of 153  4 is 1.

MIT Math Syllabus 10-3 Lesson 5: Complex numbers

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MIT Math Syllabus 10-3 Lesson 5: Complex numbers