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Let $z \neq -\iota$ be any complex number such that $\frac{z-\iota}{z+\iota}$ is a purely imaginary number. Then $z+\frac{1}{z}$ is:

(1) 0

(2) any non-zero real number other than 1.

(3) any non-zero real number.

(4) a purely imaginary number.

My Attempt:$ \frac{z- \iota}{z+\iota}$ is a purely imaginary .\

So $\frac{z- \iota}{z+\iota} = i k$ , $k \in \mathbb R$\

$\Rightarrow z = \frac{-2k+i(1-k^2)}{1+k^2}$\

We can see $\Rightarrow z\overline{z}=|z|^2 =1 $\

$\therefore z+\frac{1}{z}=z+\overline{z} = \frac{-4k}{ 1+k^2} $\

$\bullet$ if $k=1$ then $z+\frac{1}{z}=z+\overline{z} = \frac{-4k}{ 1+k^2} = -2$.\

So option $(1)$ and $(4)$is incorrect.\

$\bullet$ now we will check the range of this function $\frac{-4k}{1+k^2}$ , where $k \in \mathbb R$.\

Let's say $\frac{-4k}{1+k^2} = l$ which means \

$l+lk^2 +4k=0$\

[ As $k$ is real number , the discriminant of this quadratic equation will be non-negative]\

$\Rightarrow 16 -4l^2 \geq 0$\

$ \Rightarrow -2 \leq l \leq 2$\

So option $(2)$ , $(3)$ are incorrect.\

So no option is correct.

Can anyone please check my attempt? Have I gone wrong anywhere?

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All four options are wrong. We are given that the real part of $\frac {z-i} {z+i}$ is $0$. But $\frac {z-i} {z+i}=\frac {(z-i)(\overline {z} -i)} {|z+i|^{2}}$ so the hypothesis translates to $|z|=1$. Writing $z=e^{ia}$ where $a$ is real we see that $z+\frac 1 z=e^{ia}+e^{-ia}=2\cos a$ which can be any real number between $-2$ and $+2$.

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  • $\begingroup$ So I am right? Right? @Kavi Rama Murthy $\endgroup$ – sani Mar 30 '20 at 12:01
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    $\begingroup$ @sani Your answer is correct. I just thought my approach was simpler. $\endgroup$ – Kavi Rama Murthy Mar 30 '20 at 12:03

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