# 4.6: Trigonometric Form of Complex Numbers

**At Grade**Created by: CK-12

## Learning Objectives

- Convert complex numbers into rectangular and polar coordinates
- Use the polar form of complex numbers to explore powers of complex numbers

## Introduction

You have learned that rectangular graphs can be put into polar form, and that points in rectangular coordinates can be plotted in the polar coordinate system. In this section you will learn how to do the same process with complex numbers.

## Complex Plane

In standard form *z* = *a* + *bi*, a complex number can be graphed using rectangular coordinates (*a*, *b*). ‘a’ represents the *x* - coordinate, while ‘b’ represents the *y* - coordinate. Alternatively, the *x* - coordinate represents “real number” values, while the *y* - coordinate represents the “imaginary” values.

For example, given the complex number in standard form: *z* = 2 + 2*i*, you can graph this number in the coordinate plane To graph this point, the coordinate (2, 2) is graphed as shown below:

## Relationships among x, y, r, and θ

You can also graph complex numbers in the polar coordinate system using the familiar substitutions.

If you are given *r* and *θ*, then use

*x* = *r* cos θ and *y* = *r* sin θ

If you are given *x* and *y*, then

\begin{align*}r = \sqrt{x^2 + y^2}\end{align*}

All the same rules and procedures for converting points represented by a real pairs of numbers in the rectangular plane apply to converting complex numbers into polar form. We review the steps for conversion below.

## Trigonometric or Polar Form of a Complex Number (r cis θ)

In the example above, we graphed the complex number *z* = *a* + *bi* in rectangular coordinate system. Recall that there is another coordinate system we can use, the polar coordinate system. In a previous lesson you learned that rectangular coordinates (x, y) can be transformed into polar coordinates (*r*, *θ*).

Here we will use that basic conversion to rewrite *z* = *a* + *bi* in another (sometimes more convenient) form that is based on the polar conversion. This new form is called the **trigonometric form** of a complex number.

Start with the complex number \begin{align*}z = -1 - i\sqrt{3}\end{align*}. You can graph it in the rectangular coordinate system:

In Polar form, we find *r* with

\begin{align*}r = \sqrt{a^2 + b^2}\end{align*}

\begin{align*}= \sqrt{(-1)^2 + (-\sqrt{3})^2}\end{align*}

\begin{align*}= \sqrt{1 + 3}\end{align*}

\begin{align*}= \sqrt{4}\end{align*}

\begin{align*}= 2\end{align*}

and to find *θ*,

\begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{-\sqrt{3}} {-1}\right |\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \sqrt{3}\end{align*}

\begin{align*}\theta_{ref} = \mbox{tan}^{-1}\ \sqrt{3}\end{align*}

\begin{align*}\theta_{ref} = \frac{\pi} {3}\end{align*}

Since this angle is in the 4^{th} quadrant, \begin{align*}\theta = \frac{4\pi} {3}\end{align*}.

Now, there is a problem with using a conversion from rectangular form to polar form like

\begin{align*}a + bi \rightarrow (r, \theta)\end{align*}

\begin{align*}-1 - i\sqrt{3}\rightarrow \left (2, \frac{4\pi} {3}\right )\end{align*}

The problem is that we have lost the *i*. So, in order to “keep track” of the imaginary part, we can use another form. Recall that we used the substitutions

*x* = *r* cos *θ* and *y* = *r* sin *θ*.

Using this fact, and the values we know for *r* and *θ*, we can write

\begin{align*}z = -1 - i\sqrt{3} = 2\ \mbox{cos}\ \frac{4\pi} {3} + 2\ i\ \mbox{sin}\ \frac{4\pi} {3}\end{align*}

Finally, factoring the 2,

\begin{align*}z = 2 \left (\mbox{cos}\ \frac{4\pi} {3} + i\ \mbox{sin}\ \frac{4 \pi} {3}\right )\end{align*}

*Our new form,* \begin{align*}z = 2 \left (\mbox{cos}\ \frac{4\pi} {3} + i\ \mbox{sin}\ \frac{4\pi} {3}\right )\end{align*} is in the form z = **r**(**c**osθ + **is**inθ), where r and θ are the polar coordinates of the imaginary number. This form will be used very often in the future. It has an abbreviated form known as “rcisθ”

So, the complex number: \begin{align*}z = -1 - \sqrt{3}i\end{align*}, the rectangular point \begin{align*}(-1, -\sqrt{3})\end{align*}, the polar point: \begin{align*}\left (2, \frac{4\pi} {3}\right )\end{align*}, and \begin{align*}2 \left (\mbox{cos}\ \frac{4\pi} {3} + i\ \mbox{sin}\ \frac{4\pi} {3}\right )\end{align*} or \begin{align*}2\ \mbox{cis}\ \left (\frac{4\pi} {3}\right )\end{align*} all represent the same number.

*Each of these forms has benefits in specific situations as we will show shortly.*

## Steps for Conversion

To convert from polar to rectangular form, the distance that the point (2, 2) is from the origin can be found by

\begin{align*}d = \sqrt{x^2 + y^2}\end{align*} or \begin{align*}\sqrt{2^2 + 2^2}\ d = \sqrt{8}\end{align*} or \begin{align*}2\sqrt{2}\end{align*}

The reference angle (i.e. the corresponding angle in the first quadrant) that the line segment between the point and the origin can be found by

\begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{y} {x}\right |\end{align*}

for *z* = 2 + 2*i*,

\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{2} {2}\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = 1.\end{align*}

Since this point is in the first quadrant (both the *x* and *y* coordinate are positive) the angle must be 45^{o} or \begin{align*}\frac{\pi} {4}\end{align*} radians.

It is also possible that when tan *θ* = 1 the angle can be in the third quadrant or \begin{align*}\frac{5\pi} {4}\end{align*} radians. But this angle will not satisfy the conditions of the problem, since a third quadrant angle must have both x and y as negatives.

Note: When using \begin{align*}\mbox{tan}\ \theta = \frac{y} {x}\end{align*}, you should first consider, the quotient \begin{align*}\left |\frac{y} {x}\right |\end{align*} and find the first quadrant angle that satisfies this condition. This angle will be called the **reference angle,** denoted \begin{align*}\theta_{ref}\end{align*}. Find the actual angle by analyzing which quadrant the angle must be given the *x* and *y* signs.

The complex number 2 + 2*i* or (2, 2) in rectangular form has polar coordinates \begin{align*}\left (2\sqrt{2}, \frac{\pi} {4}\right )\end{align*}

**Example:** Find the Polar coordinates that represent the complex number \begin{align*}z = 3 - 3\sqrt{3}i\end{align*}

a = 3 and b = \begin{align*}-3\sqrt{3}\end{align*}: the rectangular coordinates of the point is \begin{align*}\left (3, -3\sqrt{3}\right )\end{align*}.

Now, draw a right triangle in standard form. Find the distance the point is from the origin and the angle the line segment that represents this distance makes with the +x axis:

We know a = 3, \begin{align*}b=-3\sqrt{3}\end{align*}

\begin{align*}r = \sqrt{3^2 + (-3\sqrt{3})^2}\end{align*}

\begin{align*}= \sqrt{9 + 27}\end{align*}

\begin{align*}= \sqrt{36}\end{align*}

\begin{align*}= 6\end{align*}

And for the angle,

\begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{(-3\sqrt{3})} {3}\right |\end{align*}

\begin{align*}\mbox{tan}\ \theta_{ref} = \sqrt{3}\end{align*}

\begin{align*}\theta_{ref} = \frac{\pi} {3}\end{align*}

But, since it is a 4^{th} quadrant angle

\begin{align*}\theta = \frac{5 \pi} {3}\end{align*}

The rectangular point \begin{align*}(3, -3\sqrt{3}i)\end{align*} is equivalent to the polar point \begin{align*}\left (6, \frac{5\pi} {3}\right )\end{align*}.

In rcisθ form, \begin{align*}(3, -3\sqrt{3}i)\end{align*} is \begin{align*}6\left (\mbox{cos}\ \frac{5\pi} {3} + i\ \mbox{sin}\ \frac{5\pi} {3}\right )\end{align*}.

## Applications, Technological Tools

Some graphing calculators can convert polar to rectangular and rectangular to polar form.

On the TI-84: go to **[ANGLE]** (or **[2nd]** function) **[APPS]**. Scroll down to 5 or “R-Pr(“ and press **[Enter]** . Next, enter the rectangular coordinates and close the parenthesis. Press **[Enter]**, the “r” value appears. Scroll down to 6R-Pθ, and the polar angle appears in decimal radian form.

Also under the **[ANGLE]** menu, commands 7 and 8 allow transformation from polar form to rectangular form.

## Lesson Summary

There are multiple ways to represent a complex number. Each way has their advantages when doing specific problems. Knowing how to convert from one representation to another gives the student flexibility in being able to work with many different problems that use complex numbers.

## Points to Consider

Complex numbers can be written in either rectangular form or in polar form. For what type of problem is it more advantageous to write a complex number in rectangular form? In polar form?

## Review Questions

- Convert the following complex number into polar form: a. \begin{align*}\sqrt{3} - i\end{align*} b. \begin{align*}9\sqrt{3} + 9i\end{align*} c. 3 - 4
*i*d. \begin{align*}\sqrt{5} - i\end{align*} - Convert the following complex numbers into rectangular form: a. 3(cos 210
^{o}+*i*sin 210^{o}) b. \begin{align*}2\left (\mbox{cos}\ \frac{\pi} {18} + i\ \mbox{sin}\ \frac{\pi} {18}\right )\end{align*} c. \begin{align*}4\ \mbox{cis}\ \left (\frac{7\pi} {4}\right )\end{align*} - Convert the rectangular/polar coordinate into polar/rectangular coordinate: a. \begin{align*}(3, 3\sqrt{3})\end{align*} b. (-2, 2) c. \begin{align*}\left (4, \frac{2\pi} {3}\right )\end{align*} d. \begin{align*}\left (-1, \frac{5\pi} {6}\right )\end{align*}

## Review Answers

- a. 2(cos 330
^{o}+*i*sin 330^{o}) b. 18(cos 30^{o}+*i*sin 30^{o}) c. 5(cos 306.9^{o}+ i sin 306.9^{o}) d. \begin{align*}\sqrt{6}(\mbox{cos} \ 335.9^\circ + i \ \mbox{sin} \ 335.9^\circ)\end{align*} - a. \begin{align*}\frac{-3\sqrt{3}} {2} - \frac{3} {2} i\end{align*} b. 1.97 + 0.35
*i*c. \begin{align*}2\sqrt{2} - 2\sqrt{2}i\end{align*} - .

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