Trigonometric Form of a Complex Number

Complex Numbers

Recall that a complex number has a real component and an imaginary component.

z = a + bi

Argand Diagram

Real axis

Imaginary axis

z = 3 – 2iz = 3 – 2i

a

bi

The absolute value of a complex number is its distance from the origin.

2 2z a b

223 2 9 4 13z

The names and letters are changing, but this sure looks familiar.

The Trig form of a Complex Number

cosa r

sinb r

2 2r a b

cos

sin

x a

r ry b

r r

The trig form of the complex number

is ( cos sin ) cos sin .

z a bi

z r ri r i

2 2

r is called the modulus and is the distance from

the origin to the point. r a b

1

is called the argument and is the angle

formed with the x-axis.

tanb

a

How is it Different?In a rectangular system, you go left or right and up or down.

In a trigonometric or polar system, you have a direction to travel and a distance to travel in that direction.

2 2z i 2 cos45 sin 45

(2,45)

z i

Polar form

Converting from Rectangular form to Trig form

2 2

1

1. Find r.

2. Find . tan

3. Fill in the blanks in cos sin

r a b

b

a

z r i

Convert z = 4 + 3i to trig form.

2 24 3 16 9

25 5

r

r

1. Find r 2. Find

1 3tan 36.9

4

3. Fill in the blanks

5 cos36.9 sin 36.9

5,36.9

z i

Polar form

Converting from Trig Form to Rectangular Form

This one’s easy.

1. Evaluate the sin and cos.

2. Distribute in rConvert 4(cos 30 + i sin 30) to rectangular form.

1. Evaluate the sin and cos 3 14

2 2i

2. Distribute the 4. 2 3 2i

Multiplying Complex Numbers

To multiply complex numbers in rectangular form, you would FOIL and convert i2 into –1.

To multiply complex numbers in trig form, you simply multiply the rs and add the thetas.

2

a bi c di

ac adi bci dbi

ac adi bci db

ac db ad bc i

1 1 1 2 2 2

1 2 1 2 1 2

cos sin cos sin

cos sin

r i r i

r r i

The formulas are scarier than it really is.

Example

1 2

1

2

2 3 2 4 cos30 sin 30

3 2 3 2 6 cos45 sin 45

multiply z z

Where z i i

z i i

1 2

2

2 3 2 3 2 3 2

6 6 6 6 6 2 6 2

6 6 6 6 6 2 6 2

6 6 6 2 6 6 6 2

z z

i i

i i i

i i

i

Rectangular form Trig form

1 2

4 cos30 sin 30 6 cos45 sin 45

4 6 cos 30 45 sin 30 45

24 cos75 sin 75

z z

i i

i

i

2 2

6 6 6 2 6 6 6 2

216 72 12 72 216 72 12 72

576 24

r

r

r

1 6 6 6 2tan 75

6 6 6 2

Dividing Complex Numbers

In rectangular form, you rationalize using the complex conjugate.

2

2 2 2

2 2

2 2 2 2

a bi

c dia bi c di

c di c di

ac adi bci bdi

c d iac adi bci bd

c dac bd bc ad

ic d c d

In trig form, you just divide the rs and subtract the theta.

1 1 1

2 2 2

11 2 1 2

2

cos sin

cos sin

cos sin

r i

r i

ri

r

Example

1

2

1

2

3 2 3 2 6 cos45 sin 45

2 3 2 4 cos30 sin 30

zdivide

z

Where z i i

z i i

Rectangular form

2

2

3 2 3 2

2 3 2

3 2 3 2 2 3 2

2 3 2 2 3 2

6 6 6 2 6 6 6 2

12 4

6 6 6 2 6 6 6 2

12 4

6 6 6 2 6 6 6 2

16 16

i

i

i i

i i

i i i

i

i i

i

Trig form

6 cos45 sin 45

4 cos30 sin 30

6cos 45 30 sin 45 30

43

cos15 sin152

i

i

i

i

2 2

6 6 6 2 6 6 6 2

16 16

216 72 12 72 216 72 12 72

256

576 9 3

256 4 2

r

r

r

1

6 6 6 2

16tan 156 6 6 2

16

De Moivre’s Theorem If is a

complex number And n is a positive integer Then

(cos sin )z r i

(cos sin )

(cos sin )

nn

n

z r i

r n i n

Who was De Moivre?

A brilliant French mathematician who was persecuted in France because of his religious beliefs. De Moivre moved to England where he tutored mathematics privately and became friends with Sir Issac Newton.

De Moivre made a breakthrough in the field of probability (writing the Doctrine of Chance), but more importantly he moved trigonometry into the field of analysis through complex numbers with De Moivre’s theorem.

But, can we prove DeMoivre’s Theorem?

(cos sin )z r i 22 (cos sin )z r i

Let’s look at some Powers of z.

2 2(cos sin )r i 2(cos sin )(cos sin )r i i 2 2 2 2(cos 2 cos sin sin )r i i 2 2 2(cos sin 2 cos sin )r i 2(cos2 sin 2 )r i

33 (cos sin )z r i

Let’s look at some more Powers of z.

2 2(cos sin ) (cos sin )r i r i 3(cos3 sin 3 )r i

44 (cos sin )z r i 3 3(cos sin ) (cos sin )r i r i

4(cos4 sin 4 )r i

It appears that:

cos i sin n cos n i sinn

Assume n=1, then the statement is true.

We can continue in the previous manor up to some arbitrary k

Let n = k, so that:

cos i sin k cosk i sink

Now find

cos i sin k1

cos i sin k1 cos k i sink cos i sin

cos i sin k1 cosk cos sink sin i cosk sin sink cos

cos i sin k1cos(k ) i sin(k )

cos i sin k1cos(k 1) i sin(k 1)

Proof:

Euler’s Formula cos sinie i

cos sinnn ii e

We can also use Euler’s formula to prove DeMoivre’s Theorem.

= ine

= cos sinn i n

So what is the use?

Find an identity for using Mr. De Moivre’s fantastic theory

cos5

cos5 i sin5 cos i sin 5

Remember the binomial expansion: 5041322314055 )()(1)()(5)()(10)()(10)()(5)()(1)( bababababababa

Apply it:

)sin)()(cos10()sin)()(cos10()sin)()(cos5())(cos1(5sin5cos 33222345 iiii

(5)(cos )(i4 sin4 ) (1)(i5 sin5 )

Cancel out the imaginery numbers:

4235 sincos5sincos10cos5cos

Now try these:

cos3

sin3

sin4

cos3 3cos sin2

3cos2 sin sin3

4 cos3 sin 4 cos sin3

Powers of Complex Numbers

This is horrible in rectangular form.

...

na bi

a bi a bi a bi a bi

The best way to expand one of these is using Pascal’s triangle and binomial expansion.

You’d need to use an i-chart to simplify.

It’s much nicer in trig form. You just raise the r to the power and multiply theta by the exponent.

cos sin

cos sinn n

z r i

z r n i n

3 3

3

5 cos20 sin 20

5 cos3 20 sin 3 20

125 cos60 sin 60

Example

z i

z i

z i

Roots of Complex Numbers

There will be as many answers as the index of the root you are looking for Square root = 2 answers Cube root = 3 answers, etc.

Answers will be spaced symmetrically around the circle You divide a full circle by the number of

answers to find out how far apart they are

The formula

cos sin

360 360 2 2cos sin cos sinn n n

z r i

k k k kz r i or r i

n n n n

k starts at 0 and goes up to n-1

This is easier than it looks.

Using DeMoivre’s Theorem we get

General Process

1. Problem must be in trig form2. Take the nth root of r. All answers

have the same value for r.3. Divide theta by n to find the first

angle.4. Divide a full circle by n to find out

how much you add to theta to get to each subsequent answer.

Example Find the 4th root of 81 cos80 sin80z i

1. Find the 4th root of 81 4 81 3r

2. Divide theta by 4 to get the first angle.

8020

4

3. Divide a full circle (360) by 4 to find out how far apart the answers are.

36090 between answers

4

4. List the 4 answers.

• The only thing that changes is the angle.

• The number of answers equals the number of roots.

1

2

3

4

3 cos20 sin 20

3 cos 20 90 sin 20 90 3 cos110 sin110

3 cos 110 90 sin 110 90 3 cos200 sin 200

3 cos 200 90 sin 200 90 3 cos290 sin 290

z i

z i i

z i i

z i i