2
$\begingroup$

can someone please help me with a trivial question?

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hans Lundmark Aug 18 '18 at 8:58
2
$\begingroup$

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 $$-3i=3e^{-i\dfrac{\pi}{2}}$$

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

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

$\endgroup$
  • $\begingroup$ Okay, sounds good. :) Thanks a lot. $\endgroup$ – numericalorange Aug 18 '18 at 21:31
2
$\begingroup$

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, $z=3e^{-i\frac{\pi}{2}}$.

$\endgroup$
  • 1
    $\begingroup$ That should be $z = 3e^{-i\frac{\pi}{2}}$ (you're missing an "$i$" in the exponent). $\endgroup$ – Deepak Aug 18 '18 at 6:52
  • $\begingroup$ exactly. thanks $\endgroup$ – Ronald Aug 18 '18 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.