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write in polar form this : $$ z=\dfrac{e^{2\theta i}-1}{e^{2\theta i}+1} $$ I use euler formula to get that $z=\tan(\theta)i$
then $\sin(\theta)=\dfrac{y}{r}$,$\cos(\theta)=\dfrac{x}{r}$ that gives
$$ z=\dfrac{\dfrac{y}{r}}{\dfrac{x}{r}}i=\dfrac{yi}{x} $$ so I did this last step $$ z=\dfrac{ye^{\dfrac{\pi}{2}i}}{xe^{2\pi i}} $$ I don't Know what to do next because it lead to last one

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    $\begingroup$ what is rectangular form? do you mean the $a + ib$ form? $\endgroup$ – samjoe Mar 31 '17 at 17:56
  • $\begingroup$ Surely you mean polar form? $\endgroup$ – John Kontol Mar 31 '17 at 17:57
  • $\begingroup$ yeah polar form, sorry I'm gonna edit that now $\endgroup$ – user11618 Mar 31 '17 at 17:59
  • $\begingroup$ No, I mean to write it in $\cos(x)+\sin(x)i$ $\endgroup$ – user11618 Mar 31 '17 at 18:02
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    $\begingroup$ Well you already have $z=i\tan(\theta)$ $\endgroup$ – samjoe Mar 31 '17 at 18:04
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If $z=i\tan(\theta)$, and if $\theta \in \mathbb{R}$, then we have $z=|z|e^{i\arg(z)}$, where

$$|z|=|\tan(\theta)|$$

and

$$\arg(z)=\begin{cases}\frac{\pi}{2}+2k\pi&,\tan(\theta)\ge 0\\\\\frac{\pi}{2}+(2k+1)\pi&,\tan(\theta)< 0\end{cases} \tag1$$

for any $k\in \mathbb{Z}$.

So, the polar form of $z$ is

$$z=|\tan(\theta)|e^{i\arg(z)}$$

where $\arg(z)$ is given by $(1)$.

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  • $\begingroup$ This answer is trivial because $e^{i\text{arg}(z)} = i*\text{sign}(\theta)$ $\endgroup$ – Cye Waldman Apr 1 '17 at 17:02
  • $\begingroup$ @CyeWaldman Yes, it is trivial. But the OP specifically requested a polar coordinate form. $i\text{sgn}(\theta)$ is Cartesian. $\endgroup$ – Mark Viola Apr 1 '17 at 17:08
  • $\begingroup$ Okay, I guess I can't argue with that. $\endgroup$ – Cye Waldman Apr 1 '17 at 17:14

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