# Cartesian and Polar Coordinate

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?

• For c, you are close, but the $z$ on the right is the same in both places. It should be $\frac \pi 4$ Feb 2, 2013 at 17:41
• A nitpick only, but please don't say things like "$\sqrt{3}=1.73$"--it causes many on this site near-physical anguish. It's okay just to say "$\sqrt{3}$" and not give the approximation, or (if you'd like) to say "$\sqrt{3}\approx 17.3$" with \approx to get that $\approx$ symbol. Feb 2, 2013 at 17:45
• @RossMillikan thanks,now i see $\cos{(-\frac{\pi}{4})}$ Feb 2, 2013 at 18:03

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

• Ok so to be sure that i understand it: d) $z_{4}=-4\cdot e^{\frac{\pi}{3}i}$ this will be $-4(\cos{\frac{\pi}{3}}+\sin{\frac{\pi}{3}})=-4(\frac{1}{2}+i\frac{\sqrt{3}}{2})=-\frac{4}{2}-\frac{4\sqrt{3}}{2}i=-2-2\sqrt{3}i$ So the Cartesian Coordinates are $(-2,-2\sqrt{3})$ $r=\sqrt{(-2)^2+(-2\sqrt{3})^2}=\sqrt{16}=4$ $\boldsymbol{\sin{\theta}}=\frac{y}{r}=-\frac{\sqrt{3}}{2}$, $\boldsymbol{\cos{\theta}}=\frac{x}{r}=-\frac{1}{2}$ so $\theta$ lies in the $3_{rd}$ Quadrant, $\theta=arctan\frac{y}{x}=arctan(\sqrt{3})$ is this correct ? Feb 2, 2013 at 19:57
• @Devid, yes, the value will be $\pi+\frac\pi3$ as the principal value of $\arctan \sqrt3$ is $\frac\pi3$ Feb 3, 2013 at 5:34
• But when i do the same with $z_{1}$ i get $arctan \frac{-1}{0}$, but this can't be. Feb 3, 2013 at 10:35
• $\phi=\begin{cases} \arctan(\frac{y}{x}) & \text{for } x>0\\ \arctan(\frac{x}{y})+\pi & \text{for} \ x<0 \\\frac{\pi}{2} & \text{for} \ x=0,y>0 \\ -\frac{\pi}{2} & \text{for} \ x=0,y<0 \end{cases}$ Feb 3, 2013 at 10:54
• en.wikipedia.org/wiki/Atan2#Definition Feb 3, 2013 at 12:38

Hint: $$e^{i\theta}=cos\theta+i\sin\theta$$

Hint: $z=(x,y)$ then $r = \sqrt {x^2+y^2}$ and $\phi=\tan^{-1}\frac{y}{x}$ and $$r e^{i\phi}= r (\cos \phi + i \sin \phi).$$

• Thanks but whats with $z_{1}$ i get $arctan(-\frac{1}{0})$ Feb 2, 2013 at 21:43
• $\phi=\frac{-\pi}{2}$
– M.H
Feb 3, 2013 at 3:20
• @MaisamHedyelloo how did you get $\frac{-\pi}{2}$ ? Feb 3, 2013 at 10:36
• ok i got it: $\phi=\begin{cases} \arctan(\frac{y}{x}) & \text{for } x>0\\ \arctan(\frac{x}{y})+\pi & \text{for} \ x<0 \\\frac{\pi}{2} & \text{for} \ x=0,y>0 \\ -\frac{\pi}{2} & \text{for} \ x=0,y<0 \end{cases}$ Feb 3, 2013 at 10:52