# How do you write the trigonometric form of 1+6i?

Oct 26, 2017

$\sqrt{37} \left(\cos {80.5}^{o} + i \cdot \sin {80.5}^{o}\right) \mathmr{and} \sqrt{37} \cdot c i s {80.5}^{o}$

#### Explanation:

Converting complex numbers into trigonometric form is very similar to converting rectangular coordinates (aka Cartesian Coordinates) into polar coordinates. There are 2 formulas that will assist (assuming the complex number is in $a + b i$ format):

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\tan {\theta}_{r e f} = | \frac{b}{a} |$

We begin by finding the value $r$:

$r = \sqrt{{1}^{2} + {6}^{2}} = \sqrt{1 + 36} = \sqrt{37}$

Next, we can determine the reference angle. For this problem, I am using degrees, though the same work could be done with radians as the measure instead.

$\tan {\theta}_{r e f} = | \frac{6}{1} |$

${\theta}_{r e f} = {80.5}^{o}$

This is only the reference angle, which will always be in the first quadrant when done this way. The complex number $1 + 6 i$ lies in the first quadrant in the complex plane, and so this angle will be sufficient for our needs.

The trigonometric form of a complex number $a + b i$ is written one of two ways:

$\left\{\begin{matrix}r \left(\cos \theta + i \cdot \sin \theta\right) & \text{Expanded Form" \\ r*cis theta & "Condensed Form}\end{matrix}\right.$

Thus, our answer is:

$\sqrt{37} \left(\cos {80.5}^{o} + i \cdot \sin {80.5}^{o}\right) \mathmr{and} \sqrt{37} \cdot c i s {80.5}^{o}$