Complex numbers can be graphed on a polar graph just like real numbers can. You will discover during this lesson that there are actually a few different ways of doing this.

### Polar Form of Complex Numbers

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.

There are three common forms of complex numbers that you will see when graphing:

- In the standard form of:
*z*=*a*+*bi*, a complex number*z*can be graphed using rectangular coordinates (*a*,*b*). ‘a’ represents the*x*- coordinate, while ‘b’ represents the*y*- coordinate. - The polar form: \begin{align*}(r, \theta)\end{align*} which we explored in a previous lesson, can also be used to graph a complex number. Recall that you can use
*x*and*y*to convert between rectangular and polar forms with: \begin{align*}r = \sqrt{x^2 + y^2}\end{align*} and \begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{y} {x}\right |\end{align*}. Unfortunately, there is a problem with using a conversion from rectangular form to polar form like:

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

or

\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.

- The third form is trigonometric form. It is often abbreviated as
, short for: z =*rcisθ***r**(**c**osθ +**is**inθ), and will be used quite often as you progress. This form comes from the substitutions:*x*=*r*cos*θ*and*y*=*r*sin*θ*.

Using this fact, and sample values of \begin{align*}2\end{align*} for *r* and \begin{align*}\frac{\pi} {3}\end{align*} for *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, we get: \begin{align*}z = 2 \left (\mbox{cos}\ \frac{4\pi} {3} + i\ \mbox{sin}\ \frac{4 \pi} {3}\right )\end{align*}

#### Summary of Forms

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.

#### 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 \begin{align*}x\end{align*} and \begin{align*}y\end{align*} 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*}

### Examples

#### Example 1

Graph in polar form: \begin{align*}z = -1 - i\sqrt{3}\end{align*}.

Here is what it looks like 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*}.

#### Example 2

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 are \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*}.

#### Example 3

Convert the following complex numbers into polar form, use a TI-84 equivalent graphing calculator:

- \begin{align*}\sqrt{3} - i\end{align*}
- \begin{align*}9\sqrt{3} + 9i\end{align*}

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.

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

#### Example 4

Plot the complex number \begin{align*}z = 12 + 9i\end{align*}.

- What is needed in order to plot this point on the polar plane?

First, we will need to know \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}.

- How could the r-value be determined?

The \begin{align*}r\end{align*} value is the hypotenuse of a triangle with two other sides, \begin{align*}A = 12\end{align*} and \begin{align*}B = 9\end{align*}. It can be determined with the Pythagorean theorem: \begin{align*}A^2 + B^2 = C^2\end{align*}.

- What is the r for this point?

The \begin{align*}r\end{align*} value for this point is \begin{align*}\sqrt{144 + 81} \to \sqrt{225} = 15\end{align*}.

- How could \begin{align*}\theta\end{align*} be determined?

\begin{align*}\theta\end{align*} can be calculated using either \begin{align*}sin \theta = \frac{9}{15}\end{align*} or \begin{align*}cos \theta = \frac{12}{15}\end{align*}.

- What is \begin{align*}\theta\end{align*} for this point?

For this point, \begin{align*}sin \theta = \frac{3}{5} \to 37^o\end{align*} or \begin{align*}cos \theta = \frac{4}{5} \to 37^o\end{align*}.

- What would \begin{align*} z = 12 + 9i\end{align*} look like on the polar plane?

\begin{align*}z = 12 + 9i\end{align*} looks like the image below when plotted on a polar plane.

#### Example 5

What quadrant does \begin{align*}z = -3 + 2i\end{align*} occur in when graphed?

The point \begin{align*}z = -3 + 2i\end{align*} occurs 3 units to the *left* and 2 units *up*, placing it in Quadrant II.

#### Example 6

What are the coordinates of z = -3 + 2i in polar form and trigonometric form?

To identify the coordinates of \begin{align*}z = -3 + 2i\end{align*} in polar form and trigonometric form:

\begin{align*}r = \sqrt{(-3^2) + (2^2)} \to \sqrt{13}\end{align*} First find \begin{align*}r\end{align*}

\begin{align*}sin \theta = \frac{2}{\sqrt{13}} \to 33.7^o\end{align*} Second, find \begin{align*}\theta\end{align*}

\begin{align*}\therefore [\sqrt{13}, 33.7^o]\end{align*} are the coordinates in polar form.

\begin{align*}\therefore r cis \sqrt{13} \left(\frac{\pi}{5}\right)\end{align*} are the coordinates in \begin{align*}r cis\end{align*} form

#### Example 7

What would be the polar coordinates of the point graphed below?

The rectangular coordinates are (4.5, 3i) therefore the complex number would be \begin{align*}z = 4.5 +3i\end{align*}

\begin{align*}r = 5.4\end{align*} Using the Pythagorean Theorem as in Q #3

\begin{align*}\theta = 33.75^o\end{align*} Using \begin{align*}sin = \frac{opp}{hyp}\end{align*} as in Q #3

\begin{align*}\therefore [5.4, 33.65^o]\end{align*} is the point in polar form \begin{align*}\therefore r cis 5.4 \left(\frac{\pi}{5}\right)\end{align*} are the coordinates in \begin{align*}r cis\end{align*} form

### Review

Plot each complex number in the complex plane. Find its polar form, \begin{align*}[r,\theta]\end{align*} and give the argument \begin{align*}\theta\end{align*} in degrees.

- a) \begin{align*}1 + i\end{align*} b) \begin{align*}i\end{align*} c) \begin{align*}(1 + i)i\end{align*}
- a) \begin{align*}-2\end{align*} b) \begin{align*}3i\end{align*} c)\begin{align*}(-2)(3i)\end{align*}
- a) \begin{align*}1 + i\end{align*} b) \begin{align*}1 - i\end{align*} c)\begin{align*}(1 + i)(1 - i)\end{align*}
- a) \begin{align*}1 + i\sqrt{3}\end{align*} b) \begin{align*}\sqrt{3} - i\end{align*} c)\begin{align*}(1 + i\sqrt{3})(\sqrt{3} - i)\end{align*}
- What are the rectangular coordinates for the point graphed below?

Compute and convert to \begin{align*}r cis\end{align*} form.

- \begin{align*}\frac{-2 - 2i}{1 - i}\end{align*}
- \begin{align*}1 + i^6\end{align*}
- \begin{align*}\frac{\sqrt{3}}{2} + \frac{1}{2}i^{10}\end{align*}

Change to polar form.

- \begin{align*}-3 -2i\end{align*}
- \begin{align*} 2\sqrt{3} - 2i\end{align*}

Change to rectangular form.

- \begin{align*}15 (cos 120^o + i sin 120^o)\end{align*}
- \begin{align*}12 \left( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right)\end{align*}
- For the complex number in standard form \begin{align*}x + iy\end{align*} find: a) Polar form b) Trigonometric form (Hint: Recall that \begin{align*} x = r cos \theta\end{align*} and \begin{align*}y = r sin \theta\end{align*})

### Review (Answers)

To see the Review answers, open this PDF file and look for section 4.8.