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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?

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    $\begingroup$ For c, you are close, but the $z$ on the right is the same in both places. It should be $\frac \pi 4$ $\endgroup$
    – Ross Millikan
    Feb 2, 2013 at 17:41
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    $\begingroup$ 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. $\endgroup$
    – Cameron Buie
    Feb 2, 2013 at 17:45
  • $\begingroup$ @RossMillikan thanks,now i see $\cos{(-\frac{\pi}{4})}$ $\endgroup$
    – Devid
    Feb 2, 2013 at 18:03

3 Answers 3

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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

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  • $\begingroup$ 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 ? $\endgroup$
    – Devid
    Feb 2, 2013 at 19:57
  • $\begingroup$ @Devid, yes, the value will be $\pi+\frac\pi3$ as the principal value of $\arctan \sqrt3$ is $\frac\pi3$ $\endgroup$
    – lab bhattacharjee
    Feb 3, 2013 at 5:34
  • $\begingroup$ But when i do the same with $z_{1}$ i get $arctan \frac{-1}{0}$, but this can't be. $\endgroup$
    – Devid
    Feb 3, 2013 at 10:35
  • $\begingroup$ $ \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} $ $\endgroup$
    – Devid
    Feb 3, 2013 at 10:54
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    $\begingroup$ en.wikipedia.org/wiki/Atan2#Definition $\endgroup$
    – lab bhattacharjee
    Feb 3, 2013 at 12:38
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Hint: $$e^{i\theta}=cos\theta+i\sin\theta$$

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

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  • $\begingroup$ Thanks but whats with $z_{1}$ i get $arctan(-\frac{1}{0})$ $\endgroup$
    – Devid
    Feb 2, 2013 at 21:43
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    $\begingroup$ $\phi=\frac{-\pi}{2}$ $\endgroup$
    – M.H
    Feb 3, 2013 at 3:20
  • $\begingroup$ @MaisamHedyelloo how did you get $\frac{-\pi}{2}$ ? $\endgroup$
    – Devid
    Feb 3, 2013 at 10:36
  • $\begingroup$ 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} $ $\endgroup$
    – Devid
    Feb 3, 2013 at 10:52

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