# 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:

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?

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 $$+2$$.