![Polar Form of a Complex
Number
We can think of complex numbers as vectors, as in
our earlier example. [See more on Vectors in 2-
Dimensions].
Our aim in this section is to write complex numbers
in terms of a distance from the origin (modulus) and
a direction (or angle/ argument) from the positive
horizontal axis.](https://image.slidesharecdn.com/anintrodcutiontocomplexnumbersjcw-151210095133/85/An-introdcution-to-complex-numbers-jcw-19-320.jpg)
![Example 2
Express −1+5i in Euler’s form (exponential form).
We need to find θ in radians (see Trigonometric Functions of Any
Angle if you need a reminder about reference angles) and r.
α=tan−1(x/y)=tan−1(1/5)≈1.37 radians
[This is 78.7∘ if we were working in degrees.]
Because our angle is in the second quadrant, we need to apply:
θ=π−1.37≈1.77
And
r=√x2+y2
=√(−1)2+(5)2
=√26
≈5.10
So −1+5i in exponential form is 5.10e1.77i](https://image.slidesharecdn.com/anintrodcutiontocomplexnumbersjcw-151210095133/85/An-introdcution-to-complex-numbers-jcw-31-320.jpg)
An introdcution to complex numbers jcw
- 1. A journey into the imaginary world Complex Numbers
- 2. Imaginary Numbers To allow for these "hidden roots", around the year 1800, the concept of √−1 was proposed and is now accepted as an extension of the real number system. The symbol used is i=√−1and i is called an imaginary number. Powers of i i=√−1 Using these, we can derive the following: i2=(√−1)2= Multiplying by i again gives us: i3=i2(i)= Continuing the process gives us: i4=i3(i)= i5=i4(i)= i6=i5(i)= What do you notice? What is i 99 i i 2 i 3 i 4
- 3. Complex Numbers Complex numbers have a real part and an imaginary part. We use z = x+iy where x and y are real Example 4: Complex numbers a. 5+6i Real part: 5, Re(5), Imaginary part: 6i, Im(6) b. −3+7i Real part: −3, Re(-3) Imaginary part: 7i, Im(7) Notation We can write the complex number 2+5i as 2+i5. There is no difference in meaning.
- 4. Basic Operations with Complex Numbers Addition and subtraction of complex numbers works in a similar way to that of adding and subtracting surds. This is not surprising, since the imaginary number i is defined as i=√−1. Addition of Complex Numbers Add real parts, add imaginary parts. Subtraction of Complex Numbers Subtract real parts, subtract imaginary parts. Example 1- Addition & Subtraction a. (6+7i)+(3−5i)= (6+3)+(7−5)i=9+2i b. (12+6i)−(4+5i)= (12−4)+(6−5)i=8+i Multiplication of Complex Numbers Expand brackets as usual, but care with i2! Example 2 - Multiplication Multiply the following. a. 5(2+7i)
- 5. Basic Operations with Complex Numbers Multiplication of Complex Numbers Expand brackets as usual, but care with i2! Example 2 - Multiplication Multiply the following. a. 5(2+7i) b. (6−i)(5i) c. (2−i)(3+i) d. (5+3i)2 e. (2√−9−3)(3√−16−1) f. (3+2i)(3−2i)
- 6. Multiplying by the conjugate 3+2i is the conjugate of 3−2i. In general: x+yi is the conjugate of x−yi and x−yi is the conjugate of x+yi. How do you explain this is words? Notice that when we multiply conjugates, our final answer is real only (it does not contain any imaginary terms). Why? We use the idea of conjugate when dividing complex numbers.
- 7. Division of Complex Numbers Earlier, we learned how to rationalise the denominator of an expression like: 5 3−√2 We multiplied numerator and denominator by the conjugate of the denominator, 3+√2: We did this so that we would be left with no radical (square root) in the denominator. Dividing with complex numbers is similar. Example 3 - Division a. Express 3−i 4−2i
- 8. More practice b. Simplify: 1−√−4 2+9i Exercises 1. Express in the form a + bi: (4+√−16)+(3−√−81) 2. Express in the form a + bi. √−4 2+√−9
- 9. Graphical Representation of Complex Numbers We can represent complex numbers in the complex plane. No – this is not the complex plane!
- 10. Graphical Representation of Complex Numbers We can represent complex numbers in the complex plane (Argand diagram). We use the horizontal axis for the real part and the vertical axis for the imaginary part. Example 1 The number 3+2i is represented by: The point A is the representation of the complex number 3+2i.
- 11. Adding Complex Numbers Graphically We can add complex numbers graphically. This is the same idea as adding vectors graphically. See more at Adding 2D Vectors. Example 2 Add 1+2i and 3−i graphically.
- 12. Adding Complex Numbers Graphically Example 2 Add 1+2i and 3−i graphically.
- 14. Modulus and argument Our aim in this section is to write complex numbers in terms of a distance from the origin (modulus) and a direction (or angle/ argument) from the positive horizontal axis. Argand Diagram
- 15. Modulus Modulus of a Complex Number The modulus of a complex number is the distance from (0,0) to P(x,y) (which represents the complex number z = x+iy) How would you find this distance 𝑧 ?
- 16. Argument We have seen the connection between complex numbers and vectors. Just like vectors we can express a complex number in terms of the angle 𝜽 between the complex number and the x-axis. We call this the argument of z or arg z. How do you find this angle? Always draw a diagram Imaginary axis z=x+iy Real axis
- 17. Argument The exam will specify if the argument is an angle between – 𝜋 < 𝜃 ≤ 𝜋 or 0 < 𝜃 ≤ 2𝜋 or given im degrees Find the modulus and argument of : 2-3i for 0 < 𝜃 ≤ 2𝜋
- 19. Polar Form of a Complex Number We can think of complex numbers as vectors, as in our earlier example. [See more on Vectors in 2- Dimensions]. Our aim in this section is to write complex numbers in terms of a distance from the origin (modulus) and a direction (or angle/ argument) from the positive horizontal axis.
- 20. Polar form • We find the real (horizontal) and imaginary (vertical) components in terms of r (the length of the vector) and θ(the angle made with the real axis): • From Pythagoras, we have: r2=x2+y2 and basic trigonometry gives us: • tan θ= • x= • y= • So we can write the polar form of a complex number as: • x+yi=r(cos θ+i sin θ) r is the absolute value (or modulus) of the complex number θ is the argument of the complex number. Or r cis θ what does this stand for? Or r ∠ θ
- 21. Example Find the polar form and represent graphically the complex number 7−5i. Express 3(cos 232∘+i sin 232∘) in rectangular (Cartesian) form. Represent 1+i√3 graphically and write it in polar form. Represent √2−i√2 graphically and write it in polar form.
- 22. Example Find the polar form and represent graphically the complex number 7−5j for 0 < 𝜃 ≤ 360 We need to find r and θ. r=√x2+y2 =√72+(−5)2 =√49+25 r=√74≈8.6 To find θ, we first find the acute angle α α=tan−1(y/x) =tan−1(5/7) ≈35.54o Now, 7−5i is in the fourth quadrant, so θ=360∘−35.54∘=324.46∘ So, expressing 7−5i in polar form, we have: 7−5i=8.6(cos 324.5∘+i sin324.5∘) We could also write this answer as 7−5i=8.6 cis 324.5∘. Also we could write: 7−5i=8.6∠324.5∘
- 23. Example Express 3(cos 232∘+i sin 232∘) in rectangular (Cartesian) form. To get the required answer, we simply multiply out the expression: 3(cos 232∘+i sin 232∘)=3 cos 232∘+i(3 sin 232∘) =−1.85−2.36i
- 24. Practice Represent 1+i√3 graphically and write it in polar form. We recognise this triangle as our 30-60 triangle from before. θ=60∘,r=2 So 1+i√3=2(cos 60∘+i sin 60∘)
- 25. Your turn Represent √2−i√2 graphically and write it in polar form. Represent graphically and give the rectangular form of 6(cos 180∘+i sin 180∘).
- 26. What did you get? Represent √2−i√2 graphically and write it in polar form. r=√x2+y2 =√(√2)2+(√2)2 =√2+2 =√4 =2 To find θ, we first find the acute angle α: α=tan−1(y/x) =tan−1(√2/√2) ≈45o The complex number is in the 4th quadrant, so θ=360∘−45∘=315∘ So we can write: √2−i√2=2(cos315∘+isin315∘)
- 27. What did you get? Represent graphically and give the rectangular form of 6(cos 180∘+isin 180∘). We can read the rectangular form of this number from the graph. 6(cos 180∘+isin 180∘)=−6
- 28. Euler’s Form (Exponential Form) of a Complex Number IMPORTANT: In this section, θ MUST be expressed in radians. We use the important constant e=2.7182818… We first met e in the section Natural logarithms (to the base e). The exponential form of a complex number is: Z = re i θ r is the absolute value (modulus) of the complex number, the same as we had before in the Polar Form; and θ is in radians
- 29. Example 1 Express 5(cos135∘+i sin 135∘) in exponential form. We have r=5 from the question. We must express θ=135∘ in radians. Recall: 1o=π/180 So 135o=135π/180 =3π/4 ≈2.36 radians So we can write 5(cos 135o+i sin135o) =5e3πi/4 ≈5e2.36i
- 30. Example 2 Express −1+5i in Euler (exponential form).
- 31. Example 2 Express −1+5i in Euler’s form (exponential form). We need to find θ in radians (see Trigonometric Functions of Any Angle if you need a reminder about reference angles) and r. α=tan−1(x/y)=tan−1(1/5)≈1.37 radians [This is 78.7∘ if we were working in degrees.] Because our angle is in the second quadrant, we need to apply: θ=π−1.37≈1.77 And r=√x2+y2 =√(−1)2+(5)2 =√26 ≈5.10 So −1+5i in exponential form is 5.10e1.77i
- 32. Rectangular form Polar form Exponential form x + yj r(cos θ+j sin θ) =r cis θ =r∠θ rejθ SUMMARY: Forms of a complex number These expressions have the same value. They are just different ways of expressing the same complex number. NOTE: NOTE: 1. For Polar Form, θ can be in degrees OR radians. 2. For Exponential Form, θ MUST be in radians.
- 33. Examples Express in Euler’s form: 1) 4.50 (cos 282.3∘+i sin 282.3∘) 2) Express in exponential form: −1−5i 3) Express in polar and rectangular forms: 2.50e3.84i
- 34. Example 2 r=√1+25 =√26 ≈5.10 θ=π+1.37=4.51 radians The graph for this example: So −1−5i=5.10 e4.51i
- 35. Example 3
- 36. Euler’s Formula- a special case! eix = cos x + i sin x Let x = What do you get? A badge from Island school! Why is this so special ?
- 37. Special case…. eiπ = cos π + i sin π eiπ = −1 + i × 0 (because cos π = −1 and sin π = 0) eiπ = −1 eiπ + 1 = 0
- 39. Solving polynomials http://www.mathsisfun.com/algebra/fundamental- theorem-algebra.html http://www.mathsisfun.com/algebra/polynomials- rule-signs.html
Editor's Notes
- #3: It cycles four different values each time multiplied by I i2=(√−1)2=−1 Multiplying by i again gives us: i3=i2(i)=−i Continuing the process gives us: i4=i3(i)=−i(i)=−(−1)=1 i5=i4(i)=1×i=i i6=i5(i)=i×i=−1 etc = i96+3 = i(4×24)+3 = i3 = –i
- #6: 10+35j 5+30j 7−j 16+30j −69−42j 13
- #8: =(7+j)/10
- #9: = (−16−13j)/85 = (6+4I)/13
- #14: https://www.youtube.com/watch?v=aP4fWMLofvo#t=53 Do orange sheet
- #21: x=r cos θ y=r sin θ
- #22: Answer next slide
- #29: We will prove this when we study calculus
- #33: http://www.intmath.com/complex-numbers/5-exponential-form.php
- #34: Change to RADIANS 4.50e4.93j
- #37: It seems absolutely magical that such a neat equation combines: e (Euler's Number) i (the unit imaginary number) π (the famous number pi that turns up in many interesting areas) 0 and 1 (also wonderful numbers!)