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## Precalculus

### Course: Precalculus > Unit 3

Lesson 6: Polar form of complex numbers# Complex number forms review

CCSS.Math:

Review the different ways in which we can represent complex numbers: rectangular, polar, and exponential forms.

## What are the different complex number forms?

Rectangular | a, plus, b, i | |

Polar | r, left parenthesis, cosine, left parenthesis, theta, right parenthesis, plus, i, sine, left parenthesis, theta, right parenthesis, right parenthesis | |

Exponential | r, dot, e, start superscript, i, theta, end superscript |

## Rectangular form

The rectangular form of a complex number is a sum of two terms: the number's start color #11accd, start text, r, e, a, l, end text, end color #11accd part and the number's start color #1fab54, start text, i, m, a, g, i, n, a, r, y, end text, end color #1fab54 part multiplied by i.

As such, it is really useful for adding and subtracting complex numbers.

We can also plot a complex number given in rectangular form in the

**complex plane**. The real and imaginary parts determine the real and imaginary coordinates of the number.*Want to learn more about complex number rectangular form? Check out this video about the complex plane and this video about adding and subtracting complex numbers.*

## Polar form

Polar form emphasizes the graphical attributes of complex numbers: start color #e07d10, start text, a, b, s, o, l, u, t, e, space, v, a, l, u, e, end text, end color #e07d10 (the distance of the number from the origin in the complex plane) and start color #aa87ff, start text, a, n, g, l, e, end text, end color #aa87ff (the angle that the number forms with the positive Real axis). These are also called start color #e07d10, start text, m, o, d, u, l, u, s, end text, end color #e07d10 and start color #aa87ff, start text, a, r, g, u, m, e, n, t, end text, end color #aa87ff.

Note that if we expand the parentheses in the polar representation, we get the number's rectangular form:

This form is really useful for multiplying and dividing complex numbers, because of their special behavior: the product of two numbers with absolute values start color #e07d10, r, start subscript, 1, end subscript, end color #e07d10 and start color #e07d10, r, start subscript, 2, end subscript, end color #e07d10 and angles start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff and start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff will have an absolute value start color #e07d10, r, start subscript, 1, end subscript, r, start subscript, 2, end subscript, end color #e07d10 and angle start color #aa87ff, theta, start subscript, 1, end subscript, plus, theta, start subscript, 2, end subscript, end color #aa87ff.

*Want to learn more about complex number polar form? Check out this video.*

## Exponential form

Exponential form uses the same attributes as polar form, start color #e07d10, start text, a, b, s, o, l, u, t, e, space, v, a, l, u, e, end text, end color #e07d10 and start color #aa87ff, start text, a, n, g, l, e, end text, end color #aa87ff. It only displays them in a different way that is more compact. For example, the multiplicative property can now be written as follows:

This form stems from Euler's expansion of the exponential function e, start superscript, z, end superscript to any complex number z. The reasoning behind it is quite advanced, but its meaning is simple: for any real number x, we define e, start superscript, i, x, end superscript to be cosine, left parenthesis, x, right parenthesis, plus, i, sine, left parenthesis, x, right parenthesis.

Using this definition, we obtain the equivalence of exponential and polar forms:

## Want to join the conversation?

- How can you calculate the arctan of e.g. 2(sqrt3)/6 without a calculator?(6 votes)
- 𝜃 = arctan(2√3∕6) ⇒

tan 𝜃 = sin 𝜃∕cos 𝜃 = 2√3∕6 = √3∕3 = 1∕√3

So, we have the ratio

sin 𝜃 : cos 𝜃 = 1 : √3, which means that if sin 𝜃 = 𝑎, then cos 𝜃 = 𝑎√3

From the Pythagorean identity we have

sin²𝜃 + cos²𝜃 = 1 ⇒ 𝑎² + 3𝑎² = 1 ⇒ 𝑎 = ±1∕2

So,

sin 𝜃 = ±1∕2, cos 𝜃 = ±√3∕2, which means that the angle we're looking for is either in the first quadrant (sin 𝜃, cos 𝜃 > 0) or the third quadrant (sin 𝜃, cos 𝜃 < 0).

Either way we're dealing with a right triangle whose legs are of lengths

1∕2 and √3∕2 and whose hypotenuse is 1.

By reflecting this triangle over its longer leg, we get an equilateral triangle (in this case all three sides are equal to 1), and thereby we know that the original triangle must be a 30-60-90 triangle.

Now we can use soh-cah-toa to find out that

sin 30° = 1∕2 and cos 30° = √3∕2

So, in the first quadrant 𝜃 = 30°, or in radians 𝜃 = 𝜋∕6,

and in the third quadrant 𝜃 = 𝜋∕6 − 𝜋 = −5𝜋∕6(21 votes)

- how do you go from 5(cos90+isin90) polar form to rectagular form?(1 vote)
- cos90 = 0

sin90 = 1

so 5(cos90+isin90) = 5 ( 0 + 1*i ) = 5i as the rectangular form.(18 votes)

- I've been trying to understand how to find ALL polar coordinates for a week and it's getting a little bit frustrating. Could someone explain to me, given polar coordinates of a point, how to find ALL the coordinates? The thing that doesn't make me understand is this 2nπ thing or adding the 2nπ. I just don't get the method....(4 votes)
- If you have a solution at π/2, then you have solutions every 2π.

π/2 + 2π is a solution, π/2 + 2π + 2π is a solution. Because rotating by 2π is a complete rotation around the circle.

So if you have

3Θ = π/4 +2πn

Θ=(1/3)(π/4 + 2πn)

Θ= π/12 + (2/3)πn

Then it is a matter of picking the right n to get the solution in the required range(6 votes)

- I think you should add one more step between being at let's say: 8cos(120) and then all of a sudden saying well that's clearly -4sqrt3 or something. It's not explained in the video either. On my calculator it just tells me -6.928 and I don't know how to get the 'exact' answer. Can you add an extra step or just allow for an answer like -6.928 to also be considered correct? Thank you so much for this free resource. I don't want to seem ungrateful I just don't get how to do that part!(5 votes)
- cos(120 degrees) = cos(2pi/3) = -sqrt(3)/2

8*cos(120 degrees) = 8 * -sqrt(3)/2 = -4sqrt(3)

(Exact trig values are found on the unit circle)(1 vote)

- How is A complex number question such as z=10 and theta = 210 degrees answered with a square root? it says finding a is: a = 10cos210degrees

and the answer is:

a=-5sqrt3

Well, why must the answer be -5sqrt3 and not -8.660254038 rounded to -8.660

We feel we are missing something somewhere to not know this.

Help, please.(3 votes)- They using special triangles(check trig section) to determine the answer. It looks like they wanted the exact answer.(4 votes)

- What is the use of the exponential form i didn't understand it ?(2 votes)
- It's more compact and easier to write, but I'm guessing we'd have to get into Euler's sequences in calculus to get why, exactly. I don't understand, either.(5 votes)

- Why the product of two numbers with angles theta1 and theta2 will have an angle theta1+theta2 ?

and Why the argument of z^n = n*theta ?

Thanks!(4 votes) - For the exercise on polar and rectangular forms of complex numbers, can we use a calculator or is there a way to do the calculations by hand?(2 votes)
- There is no simple way to evaluate trigonometric functions by hand for arbitrary angles. It is only feasible to use a calculator for most angles.(4 votes)

- Can we take out polar form of pure imaginary numbers(2 votes)
- Yes. The polar form of an imaginary number X×i equals Xcis(π/2).(3 votes)

- do you use theta in degrees or radians?(2 votes)
- All angles shall be in radians unless indication of a different unit is used. (For e^iθ, it is in radians.)(2 votes)