# Polar form from a complex number?

How do I solve for the polar form of $$-3\sqrt{2} - 3\sqrt{3}\,i\; ?$$

I think I solved for $$r$$ which is $$3\sqrt{5}$$ from using $$r=\sqrt{a^2+b^2}.$$
When I look for theta I use $$\tan\theta = \frac ab,$$ but when doing so I get $$\frac{\sqrt{6}}{ 2}.$$ Anyway I looked in the answer key for $$\theta$$ and it's in degrees. Around $$230.8^{\circ}$$. How do I figure that out from $$\frac{\sqrt{6}}{ 2}?$$ Did I even do it right ?

• The range of the arctangent is $[-\pi/2,\pi,2]$ and does not cover the plane. One needs to account for the signs of $a$ and $b$ to recover the correct value of $\theta$. In this case $\theta =\pi +\arctan(\sqrt 6/2)$. Apr 14, 2018 at 3:15
• You have the right idea. Note that the sign of $\cos \theta$ and $i\sin \theta$ is negative, $\theta$ is in QIII. Now pull out your calculator and get $\arctan \frac {\sqrt 6}{2} = 50.7$ and add 180 to find the corresponding angle in the appropriate quadrant. Apr 14, 2018 at 3:24

Same way always:

$z = -3\sqrt{2} + 3\sqrt 3 i = re^{i\theta}$ where $r = |-3\sqrt{2} + 3\sqrt 3 i| = \sqrt{(-3\sqrt{2})^2 + (3\sqrt 3)^2} = \sqrt{9*2 + 9*3}=\sqrt {45} = 3\sqrt 5$.

And $\theta$ is so that $\cos \theta = \frac {-3\sqrt{2}}r=\frac {-3\sqrt{2}}{3\sqrt 5}=\frac {-\sqrt 2}{\sqrt 5}; \sin\theta = \frac {3\sqrt{3}}r= \frac {3\sqrt{3}}{3\sqrt 5}=\frac {\sqrt 3}{\sqrt 5}; \tan \theta = \frac {3\sqrt{3}}{-3\sqrt {2}}= -\frac{\sqrt{3}}{\sqrt{2}}$.

As $\cos \theta <0$ and $\sin \theta > 0$ we know $\frac \pi 2 < \theta < \pi$.

$\theta = \arctan \frac {-\sqrt 2}{\sqrt 5} = 2.45687$

So $z = 3\sqrt 5e^{2.45687i}$.

A complex number $z$ can be written as: $$z=a+bi,\,a,b\in\mathbb{R}$$ To convert it to its polar form, one must obtain the magnitude and angle first. The magnitude can be resolved by using Pythagoras' theorem like so. $$|z|=\sqrt{\Re^2(z)+\Im^2(z)}=\sqrt{a^2+b^2}$$ Calculating the angle can be obtained through trigonometry (multiples of $\tau$ may be added due to the periodic nature of the angle): $$\theta=\operatorname{atan2}(\Im(z),\Re(z))+\tau n=\operatorname{atan2}(b,a)+\tau n,\,n\in\mathbb{Z}$$ The polar form of $z$ is denoted as $|z|\cdot e^{\theta i}$, which can be expanded using Euler's Formula to $|z|\left(\cos(\theta)+\sin(\theta)i\right)$. You provided the complex number (rearranged): $$z=\left(-3\sqrt{2}\right)+\left(-3\sqrt{2}\right)i$$ $$|z|=3\sqrt{5},\,\theta\approx4.08$$ Substituting in the values to the polar form can approximate $z$: $$z\approx3\sqrt{5}\cdot e^{4.08i}$$