# 5.3.7: Polar Form of a Complex Number

- Page ID
- 14909

Conversion of \(a + bi\) to \((a, b)\), \((r, theta)\), and \(r\; cis \;theta\).

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.

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

\(a+bi\rightarrow (r,\theta )\)

or

\(−1−i\sqrt{3}\rightarrow \left(2,\dfrac{4\pi }{3}\right)\)

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 \(r\; cis \;\theta \), short for: \(z =(\cos \theta +**r**\sin \theta )\), and will be used quite often as you progress. This form comes from the substitutions: \(x = r \cos \theta \) and \(y = r \sin \theta \).**i**

using this fact, and sample values of 2 for \(r\) and \(\dfrac{\pi }{3}\) for \(\theta \), we can write

\(z=−1−i\sqrt{3}=2 \cos \dfrac{4 \pi }{3}+2 i \sin \dfrac{4 \pi }{3}\)

Finally, factoring the 2, we get: \(z=2 \left(\cos \dfrac{4 \pi }{3}+i \sin \dfrac{4 \pi }{3}\right)\)

**Summary of Forms**

The complex number: \(z=−1−\sqrt{3}i\), the rectangular point \((−1,−\sqrt{3})\), the polar point: \(\left(2,\dfrac{4 \pi }{3}\right)\), and \(2\left(\cos \dfrac{4 \pi }{3}+i \sin \dfrac{4 \pi }{3}\right)\) or \(2 \; cis \; \left(\dfrac{4 \pi }{3}\right)\) 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

\(d=\sqrt{x^2+y^2} \text{ or } \sqrt{2^2+2^2} \qquad d=\sqrt{8} \text{ or } 2\sqrt{2}\)

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

\(\tan \theta_{ref}=\left| \dfrac{y}{x} \right|\)

for \(z = 2 + 2i\),

\(\begin{aligned} \tan \theta_{ref}=22 \\ \tan \theta_{ref}=1\end{aligned}\)

Since this point is in the first quadrant (both the * x* and

*coordinate are positive) the angle must be \(45^{\circ}\) or \(\dfrac{\pi }{4}\) radians.*

*y*It is also possible that when \(\tan \theta = 1\) the angle can be in the third quadrant or 5\pi 4 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 \(\tan \theta =\dfrac{y}{x}\), you should first consider, the quotient \(\left|yx \right|\) and find the first quadrant angle that satisfies this condition. This angle will be called the ** reference angle,** denoted \(\theta_{ref}\). Find the actual angle by analyzing which quadrant the angle must be given the

*and*

*x**signs.*

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

Example \(\PageIndex{1}\)

Graph in polar form: \(z=−1−i\sqrt{3}\).

**Solution**

Here is what it looks like in the rectangular coordinate system:

In polar form, we find \(r\) with

\(\begin{aligned} r&=\sqrt{a^2+b^2}\\ &=\sqrt{(−1)^2+(−\sqrt{3})^2}\\ &=\sqrt{1+3}\\\ &=\sqrt{4} \\ &=2 \end{aligned} \)

and to find \(\theta \),

\(\begin{aligned} \tan \theta_{ref}&=\left|\dfrac{−\sqrt{3}}{−1}\right| \\ \tan \theta_{ref}&=\sqrt{3} \\ \theta_{ref}&=tan−1 \sqrt{3}\\ \theta_{ref}&=\dfrac{\pi }{3} \end{aligned}\)

Since this angle is in the 4^{th} quadrant, \(\theta =\dfrac{4 \pi }{3}\).

Example \(\PageIndex{2}\)

Find the polar coordinates that represent the complex number \(z=3−3\sqrt{3}i\).

**Solution**

\(a = 3\) and \(b = −3\sqrt{3}\): the rectangular coordinates of the point are \((3,−3\sqrt{3})\).

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 \(a = 3\), \(b=−3\sqrt{3}\)

\(\begin{aligned} r&=\sqrt{3^2+(−3\sqrt{3})^2}\\ &=\sqrt{9+27}\\&=\sqrt{36} \\ &=6 \end{aligned}\)

And for the angle,

\(\begin{aligned} \tan \theta_{ref}&=\left|\dfrac{(−3\sqrt{3})}{3}\right| \\ \tan \theta_{ref}&=\sqrt{3} \\ \theta_{ref}&=\dfrac{\pi }{3}\end{aligned}\)

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

\(\theta =\dfrac{5 \pi }{3}\)

The rectangular point \((3,−3\sqrt{3}i)\) is equivalent to the polar point \(\left(6,5\dfrac{\pi }{3}\right)\).

In \(r\; cis \;\theta\) form, \((3,−3\sqrt{3}i)\) is \(6\left(\cos \dfrac{5 \pi }{3}+i \sin \dfrac{5 \pi }{3}\right)\).

Example \(\PageIndex{3}\)

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

- \(\sqrt{3}−i\)
- \(9\sqrt{3}+9i\)

**Solution**

On the TI-84: go to ** [ANGLE]** (or

**function)**

**[2nd]****. Scroll down to 5 or “R-Pr(“ and press**

**[APPS]****. Next, enter the rectangular coordinates and close the parenthesis. Press**

**[Enter]****, the “r” value appears. Scroll down to 6R-P\theta , and the polar angle appears in decimal radian form.**

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

Example \(\PageIndex{4}\)

Plot the complex number \(z=12+9i\).

**Solution**

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

First, we will need to know \(r\) and \(\theta \).

- How could the r-value be determined?

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

- What is the \(r\) for this point?

The \(r\) value for this point is \(\sqrt{144+81}\rightarrow \sqrt{225}=15\).

- How could \(\theta\) be determined?

\theta can be calculated using either \(\sin \theta =\dfrac{9}{15}\) or \(\cos \theta =\dfrac{12}{15}\).

- What is \(\theta \) for this point?

For this point, \(\sin \theta =\dfrac{3}{5}\rightarrow 37^{\circ}\) or \(\cos \theta =45\rightarrow 37^{\circ}\).

- What would \(z=12+9i\) look like on the polar plane?

\(z=12+9i\) looks like the image below when plotted on a polar plane.

Example \(\PageIndex{5}\)

What quadrant does \(z=−3+2i\) occur in when graphed?

**Solution**

The point \(z=−3+2i\) occurs 3 units to the * left* and 2 units

*, placing it in Quadrant II.*

*up*Example \(\PageIndex{6}\)

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

**Solution**

To identify the coordinates of \(z=−3+2i\) in polar form and trigonometric form:

\(\begin{aligned} r&=\sqrt{(−3^2)+(2^2)}\rightarrow \sqrt{13} && \text{First find } r \\ \sin \theta &=2\sqrt{13} \rightarrow 33.7^{\circ} && \text{Second, find }\theta \\ &\therefore [\sqrt{13} ,33.7^{\circ}] && \text{are the coordinates in polar form.} \\ &\therefore r\; cis \;\sqrt{13} \left(\dfrac{\pi}{5}\right) && \text{are the coordinates in } r\; cis \; \text{ form} \end{aligned}\)

Example \(\PageIndex{7}\)

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

**Solution**

The rectangular coordinates are \((4.5, 3i)\) therefore the complex number would be \(z=4.5+3i\)

\(r=5.4\) using the Pythagorean Theorem as in Q #3

\(\theta =33.75^{\circ}\) using \(\sin =\dfrac{\text{opp}}{\text{hyp}}\) as in Q #3

\(\therefore [5.4,33.65^{\circ}]\) is the point in polar form \(\therefore r\; cis \;5.4\left(\dfrac{\pi}{5}\right)\) are the coordinates in \(r\; cis \;\) form

**Review**

Plot each complex number in the complex plane. Find its polar form, [r,\theta ] and give the argument \theta in degrees.

- a) \(1+i\) b) \(i\) c) \((1+i)i\)
- a) \(−2\) b) \(3i\) c) \((−2)(3i)\)
- a) \(1+i\) b) \(1−i\) c) \((1+i)(1−i)\)
- a) \(1+i\sqrt{3}\) b) \(\sqrt{3}−i c)(1+i\sqrt{3})(\sqrt{3}−i)\)
- What are the rectangular coordinates for the point graphed below.

Compute and convert to \(r\; cis \;\) form.

- \(\dfrac{−2−2i}{1−i}\)
- \(1+i^6\)
- \(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i^{10}\)

Change to polar form.

- \(−3−2i\)
- \(2\sqrt{3}−2i\)

Change to rectangular form.

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

**Review (Answers)**

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

## Vocabulary

Term | Definition |
---|---|

\(r\; cis \;\theta\) | \(r\; cis \;\theta \) is shorthand for the expression \(r\cos \theta +ri\sin \theta \). |

complex number |
A complex number is the sum of a real number and an imaginary number, written in the form a+bi. |

polar coordinate system |
The polar coordinate system is a special coordinate system in which the location of each point is determined by its distance from the pole and its angle with respect to the polar axis. |

polar form |
The polar form of a point or a curve is given in terms of r and \(\theta \) and is graphed on the polar plane. |

quadrant |
A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise. |

Reference Angle |
A reference angle is the angle formed between the terminal side of the angle and the closest of either the positive or negative x-axis. |

trigonometric form |
To write a complex number in trigonometric form means to write it in the form \(r\cos \theta +ri\sin \theta \). \(r\; cis \;\theta \) is shorthand for this expression. |

trigonometric polar form |
To write a complex number in trigonometric form means to write it in the form \(r\cos \theta +ri\sin \theta \). \(r\; cis \;\theta \) is shorthand for this expression. |

## Additional Resources

Video: Complex Numbers in Trigonometric Form

Practice: Polar Form of a Complex Number