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I have trouble doing this question:
$w=kz/(z^2+1)$ where $z^2$ is not equal to $-1$ $Im(w)=Im(k)=0$ and $Im(z)$ is not equal to $0$, prove that modulus of $z$ is $1$.

I don't even know how I should start, so if you could give me any hint I would be really thankful.

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  • $\begingroup$ Hint: Since $w$ is real, so set $w=\bar{w}$ and simplify. $\endgroup$
    – Anurag A
    May 29 '17 at 17:47
  • $\begingroup$ You also need the condition $\,k \ne 0\,$. $\endgroup$
    – dxiv
    May 29 '17 at 17:55
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Hint: $$ 0 = 2i \operatorname{Im}(w)=w-\bar w=\frac{kz}{z^2+1}-\frac{k \bar z }{\bar z^2+1}=\frac{k(z \bar z^2 + z - z^2 \bar z -\bar z)}{(z^2+1)(\bar z^2+1)}=\frac{k(z-\bar z)(1-|z|^2)}{|z^2+1|^2} $$

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Let $z=r(\cos\theta+i\sin\theta)$. Note that

$$z+\frac{1}{z}=\frac{k}{w}$$

is real.

So we have

$$r(\cos\theta+i\sin\theta)+\frac{1}{r}(\cos\theta-i\sin\theta)$$

is real.

Therefore,

$$\left(r-\frac{1}{r}\right)\sin\theta=0$$

Since $\text{Im}(z)\ne0$, $\sin\theta\ne0$.

So $r=1$.

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