# Write $-3i$ in polar coordinates.

Write $-3i$ in polar coordinates.

So $z=a+bi=rcis\theta$ with $r=\sqrt{a^{2}+b^{2}}$ and $\theta = arctan\frac{b}{a}$. However, what if $a=0$ such as the case for $-3i$? I am confused!

• Warning: the formula $\theta = \arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead. – Hans Lundmark Aug 18 '18 at 8:58

According to a definition we always have $$-\pi<\theta\le \pi$$here we can write $$\theta=\tan^{-1}\dfrac{-3}{0}=-\dfrac{\pi}{2}\\r=3$$therefore $$-3i=3e^{-i\dfrac{\pi}{2}}$$

• Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $\tan^{-1}\dfrac{-3}{x}$ gets to $\dfrac{-\pi}{2}$ – Truth-seek Aug 18 '18 at 7:24
• Sure! All of these discussions are true in limit.... – Mostafa Ayaz Aug 18 '18 at 8:55
• I love this idea of using limits! Thanks for the insightful response! – numericalorange Aug 18 '18 at 21:31
• You're welcome ^____^ – Mostafa Ayaz Aug 19 '18 at 20:05

Try to draw a picture. I think you will be able to see the angle:

• Okay, sounds good. :) Thanks a lot. – numericalorange Aug 18 '18 at 21:31

Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^{i\theta}$ where $r=\sqrt{x^2+y^2}=3$ and the angle is $-\frac{\pi}{2}$. i.e, $z=3e^{-i\frac{\pi}{2}}$.

• That should be $z = 3e^{-i\frac{\pi}{2}}$ (you're missing an "$i$" in the exponent). – Deepak Aug 18 '18 at 6:52
• exactly. thanks – Ronald Aug 18 '18 at 6:53