0
$\begingroup$

This is an excerpt from my textbook:

Find the argument and expression in polar coordinates for (a) $z=3i$ and (b) $w=-\sqrt3+i$

(a) $|-3i| = |-3||i| = 3$, so, in polar coordinates, $-3i = 3(\cos\theta +i\sin\theta)$, and thus $\cos\theta=0$ and $\sin\theta=-1$. This yields that $\arg(-3i)=\theta=\frac{3\pi}2+2\pi k,\;k\in\mathbb Z$ Thus, in polar coordinates, $-3i = 3e^{i3\pi/2}$

(b) $|-\sqrt3+i| = \sqrt{3+1} = 2$, so in polar coordinates, $-\sqrt3+i=2(\cos\theta+i\sin\theta)$, thus $\cos\theta = -\frac{sqrt3}2$ and $\sin\theta=\frac12$. Therefore, $\arg(-\sqrt3+i)=\theta=\frac{5\pi}6+2\pi k,\;k\in\mathbb Z$. Thus, in polar coordinates, $-\sqrt3+i=2e^{i5\pi/6}$

What I do not understand is quite simple I think: how does he go from $\cos\theta=0$ and $\sin\theta=-1$ to $\arg(-3i)=\frac{3\pi}2 +2\pi k$ And please explain it for the second part also. I just dont seem to understand the unit circle properly.

$\endgroup$
0
$\begingroup$

For the unit circle, each time you reach another axis the sine and cosine of an angle change from negative to positive (and vice versa). Finding $\arg (a+bi)$ boils down to which quadrant you're in to get the correct angle.

Thus, using definition of $\arg (a+bi) = \arctan \frac {b}{a}$, and that $k \in \mathbb Z$...

(a) $-3i \rightarrow 0-3i$, so in this case $a=0$ and $b=-3$. Thus $\arg (0-3i) \rightarrow \arctan \frac {-3}{0}$, which only occurs when $\theta = \frac {3\pi}{2}+2 \pi k$. (In this case, $\tan \theta$ is undefined for $\frac {(2k+1)\pi}{2}$; we use $3\pi /2$ as $\sin \theta$ is negative and $\cos \theta$ is $0$ in the 3rd quadrant.) Since $|0-3i| = 3$, we can then write $-3i$ as $3e^{3i\pi/2}$.

(b) ${-\sqrt 3 +i}$ yields $a=-\sqrt{3}$ and $b=1$, so $\arg ({-\sqrt 3 +i}) = \arctan (-\frac {\sqrt{3}}{3})$, which only occurs when $\theta = \frac {5\pi}{6}+2 \pi k$. ($\tan \theta$ is negative in the second and fourth quadrants.) Since $|{-\sqrt 3 +i}| = 2$, we can then write ${-\sqrt 3 +i}$ as $2e^{5i\pi/6}$.

$\endgroup$
1
  • $\begingroup$ Thank you very much, its much clearer now :) $\endgroup$ – uhakdt Oct 14 '18 at 0:40

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.