# Complex Numbers

### Written by tutor Colin D.

## How to Envision Complex Numbers Graphically: The Complex Plane

- The complex number x + yi corresponds to the point with coordinates (x, y)
- The x-axis is the
*real axis* - The y-axis is the
*imaginary axis* - Real numbers are associated with points on the x-axis

For example: x = x + 0i <- -=""> (x,0) - Imaginary numbers are associated with points on the y-axis

For example: yi = 0 + yi <- -=""> (0,y)

## How to Find a Point (P) in the Complex Plane

- Any point in the complex plane can be identified by the coordinate pair (r, θ)
- r = distance from the origin to point P (i.e., line segment OP)
- θ = angle from the positive x-axis (between Quadrants I and IV) to segment OP
- All points on the terminal side can be expressed as (r cos θ, r sin θ)

-Because cos θ = adjacent/hypotenuse, and hypotenuse = r, to solve for θ, one would proceed: cos θ = x/r.

Solving this for x would result in x = r cos θ

-Likewise, because sin θ = opposite/hypotenuse, solving for θ would result in sin θ = y/r.

Solving this for y would result in y = r sin θ - Piecing it altogether:

-If we have complex number x + yi

-Then P has coordinates (x,y)

-And x = r cos θ, y = r sin θ

## Trigonometric (Polar) Form

- The trigonometric form of “x + yi” is r(cos θ + i sin θ)

-This can be derived from earlier equivalences. Because when we had x + yi, we found x = r cos θ and

y = r sin θ, we can replace x and y with r cos θ and r sin θ, respectively:

**x + yi = (r cos θ) + (r sin θ)i**

-By factoring out the "r" and multiplying by the "i," this turns into:

**r(cos θ + i sin θ)** **r = Modulus or Absolute Value**

r = (x^{2}+ y^{2})^{1/2}

r =*must be NON-negative***θ = Argument**of the complex number

-Any angle coterminal with θ is also an argument for the same complex number

tan θ = y/x -> θ = arc tan (y/x)

## Rectangular (Standard) Form

- Rectangular form is "x + yi"

## How to Change from Rectangular Form to Trigonometric Form

- If A = 2 + 2i

-First, find "r." Remember, "r" - called the modulus - is the absolute value of the hypotenuse formed by

sides "x" and "y"

r = √(2^{2}+ 2^{2})

r = √(4+4)

r = √8

r = 2√2

-Next, find θ. Remember, θ is called the argument, and is found through the following equation:

tan θ = y/x because the tangent of the angle formed by "r" and the "x-axis" equals the opposite side

divided by the adjacent side (i.e., the y-value divided by the x-value).

θ = arc tan (y/x)

θ = arc tan (2/2)

θ = arc tan (1)

θ = 45° - Then trigonometric form is found by plugging in "r" and "θ":

A = r(cos θ + i sin θ)

A = 2√2(cos 45° + i sin 45°)

## How to Change from Trigonometric Form to Rectangular Form

- If B = 3√3 (cos 330° + i sin 330°)

r = 3√3

cos 330° = √(3/2)

sin 330° = -1/2

- Then 3√3 (√(3/2) + -1/2 i) -> 9/2 - i(3√3)/2

## How to Express Complex Numbers in *Proper* Trigonometric Form

- Always remember a few essentials about proper trigonometric form:

-The modulus (r) must always be non-negative

It is the absolute value of the diagonal from the point itself to the origin.

-The parenthetical expression must be of the form: cos θ + i sin θ.

Make sure each term is written as a positive amount. - Example: z = 2(cos 30° - i sin 30°)

-First, express z in rectangular form:

2(√(3/2) - 1/2 i) -> √3 - 1i

-Thus, on a graph, this would consist of moving √3 units to the right, 1 unit down, resulting in a point in

Quadrant IV.

r = √((√3)^{2}+ 1^{2}) -> √4 -> √2

Using tan θ = y/x, we derive:

tan θ = 1/√3 -> arctan 1/√3 = -30° -> hence, θ = -30°

-Finally, substitute:

z = 2[cos (-30°) + i sin (-30°)]

## Multiplication & Division in Trigonometric Form

- NOTE: While
*rectangular*form makes addition/subtraction of complex numbers easier to conceive of,*trigonometric*form is the best method of conceiving of complex for multiplication/division purposes. - If you intend to
**multiply**two complex numbers, z1 = r1 (cos θ1 + i sin θ1), and z2 = r2 (cos θ2 + i sin θ2), the product is derivable by following a few simple steps:

-**Multiply the moduli**to find the product modulus: r1 times r2

-**Add the arguments**to find the**sum argument**: cos (θ1 + θ2) + i sin (θ1 + θ2)

-Multiply the product modulus by the sum argument: r1r2 [cos (θ1 + θ2) + i sin (θ1 + θ2)] - To divide two complex numbers:

-**Divide the moduli**to get the**quotient modulus**: r1/r2

-**Subtract the arguments**to get the**difference argument**: cos (θ1 – θ2) + i sin (θ1 – θ2)

-Multiply the quotient modulus by the difference argument: r1/r2 [cos (θ1 – θ2) + i sin (θ1 – θ2)] - Example:

z1 = √(3/2 + (1/2)i

z2 = -2 – 2i

Find z1 * z2:

(1) Express each in trigonometric form

z1 = 2(cos 30° + i sin 30°)

z2 = 2√2(cos 225° + i sin 225°)

(2) Multiply moduli:

2 * 2√2 = 4√2

(3) Add arguments:

cos(30° + 225°) + i (sin 30° + 225°)

(4) Triangular Form = 4√2 [cos(30° + 225°) + i (sin 30° + 225°)]

4√2[cos (255°) + i (sin 255°)]

To find in Rectangular Form, evaluate the cos 255° and sin 255° and simplify:

4√2 [cos (255°) + i (sin 255°)]

With the sum and difference formulae:

cos (a+b) = cos a cos b – sin a sin b

sin (a+b) = sin a cos b + sin b cos a

With calculator:

cos 255° = -.2588

sin 255° = -.9659

-1.464 – 5.464i

## DeMoiver's Theorem

- By repeating the multiplication procedure outlined just above, one may derive DeMoivre’s Theorem, which allows us to compute powers and roots of complex numbers.
- To illustrate, if we were to continue to multiply z = r (cos θ + i sin θ) by itself, we’d get:

z^{2}= r^{2}(cos 2θ + i sin 2θ)

z^{3}= r^{3}(cos 3θ + i sin 3θ)

z^{4}= r^{4}(cos 4θ + i sin 4θ) - For negative exponents, it unfolds in the following pattern:

z^{-1}= r^{-1}[(cos(-θ)) + i sin (-θ)]

z^{-2}= r^{-2}[(cos(-2θ)) + i sin (-2θ)] - Formally stated as a rule, DeMoivre’s Theorem reads:

z^{n}= r^{n}(cos nθ + i sin nθ) - EXAMPLE:

(1 + √3i)^{5}

-In trigonometric form:

2(cos 60° + i sin 60°)

-Apply DeMoivre's Theorem:

2^{5}[cos 5(60°) + i sin 5(60°)]

32 (cos 300° + i sin 300°)

32 (1/2 + i(-√3)/2)

16 – (16√3)i

## Roots of Complex Numbers

- Some basics about visualizing the roots of complex numbers:

-The*n*roots of a complex number all lie on the circle formed within the complex plane with center at the origin and radius = (r)^{(1/n)}

-The*n*roots on said circle are all equally spaced, beginning at K = 0 and proceeding until k = n-1, progressing at arguments (i.e., intervals) differing by 360°/*n* - Formula:

-Given any positive integer, n, then the nonzero complex number z (where z = r (cos θ + i sin θ)) has exactly*n*distinct*n*th roots, given by the following equation, in which k = 0, 1, 2,...., (n-1):

W (sub k) = (r)^n [cos (θ/n + k * 360°/n) + i sin (θ/n + k * 360°/n)] - Example: Find the 6th roots of 5 + 12i

(1) Write 5 + 12i in trigonometric form:

r = √(5^{2}+ 12^{2}) = 13

θ = arctan (12/5) ~ 67.38°

Thus, 5 + 12i = 13(cos θ + i sin θ)

(2) Since we're looking for sixth roots (*n*= 6), we replace*n*with 6 and simplify:

W (sub k) = 13^{1/6}[cos(θ/6 + k * 360/6) + i sin (θ/6 + k * 360/6)]

W (sub k) = 1.533 [cos (67.38/6 + k * 60) + i sin 11.23 + 60k)]

W (sub k) = 1.533 [cos(11.23 + 60k) + i sin (11.23 + 60k)]

(3) Plug in the various values of k up to (n-1) where*n*= 6 (because of the sixth roots).

K = 0 -> 1.504 + .299i

K = 1 -> .493 + 1.451i

K = 2 -> -1.01 + 1.153i

K = 3 -> -1.504 + -.299i

K = 4 -> -.493 + -1.451i

K = 5 -> 1.01 + -1.153i

Sign up for free
to access more trigonometry resources like . WyzAnt Resources features blogs, videos, lessons, and more about trigonometry and over 250 other subjects. Stop struggling and start learning today with thousands of free resources!

# Need more help with trigonometry?

Sign up for free to access thousands of educational resources from expert tutors. Stop struggling and start learning today!