Unit circle method for converting complex number into polar coordinate

Question

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.

Answer 1

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 $2e^{5i\pi/6}$.