I should give the Cartesian Coordinates $(x,y)\in \mathbb{R\times R}$ and Polar Coordinates $(r,\varphi)\in R^+\times [0,2\pi)$ of the following Complex Numbers:
a) $z_{1}=-i$
b) $z_{2}=\sqrt{3}+i$
c) $z_{3}=3\sqrt{2}\cdot e^{- \frac{\pi}{4}i}$
d) $z_{4}=-4e^{\frac{\pi}{3}i}$
Can someone help me solve this. I found the Cartesian coordinates of a) $(0,-1)$ and b) $(\sqrt{3} \approx1.73, 1)$ but what are the Cartesian coordinates of $z_{3},z_{4}$ and what should i do to find the Polar Coordinates ?
I just got c) i think. I must use the Euler Formula ${ e }^{ iz }=\cos { z+i\sin { z } }$ so it will be $3\sqrt{2}(\cos { (0) } +i\sin { (-\frac { \pi }{ 4 } } )$ right?
We know if $z$ is $(x,y)$ in the Cartesian Coordinates and $(r,\theta)$ in the Polar Coordinates,
$x=r\cos\theta$ and $y=r\sin\theta$ where $r$ is conventionally taken as non-negative
So, $x^2+y^2=r^2\implies r=+\sqrt{x^2+y^2}$
and $\tan\theta =\frac yx\implies \theta=\arctan \frac yx,$ the quandrant of $\theta$ will be dictated by the signs of $\sin\theta$ and $\cos\theta$
For the last two cases, we also need Euler Formula or Identity.
For $(c),3\sqrt2e^{-\frac\pi4i}=3\sqrt2(\cos(-\frac\pi4)+i\sin(-\frac\pi4))$ $=3\sqrt2(\frac1{\sqrt2}-i \frac1{\sqrt2})=3-i$
So, $r=\sqrt{3^2+1^2}=\sqrt{10},\sin\theta= -\frac1{\sqrt{10}}<0,\cos\theta=\frac3{\sqrt{10}}>0$ so $\theta$ lies in the 4th Quadrant.
So, $\theta=\arctan\left(-\frac13\right)$ which lies in the 4th Quadrant