# writing in polar form (complex numbers)

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

• what is rectangular form? do you mean the $a + ib$ form? – samjoe Mar 31 '17 at 17:56
• Surely you mean polar form? – John Kontol Mar 31 '17 at 17:57
• yeah polar form, sorry I'm gonna edit that now – user11618 Mar 31 '17 at 17:59
• No, I mean to write it in $\cos(x)+\sin(x)i$ – user11618 Mar 31 '17 at 18:02
• Well you already have $z=i\tan(\theta)$ – samjoe Mar 31 '17 at 18:04

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 $(1)$.

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