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Let $w = \dfrac{z+i}{z-i}$, then $w^{2016} = 1$, hence we solve for $w$ first. $$|w^{2016}| = |1+ 0i| = 1 \Rightarrow |w| = 1$$ $$\text{arg}(w^{2016}) = 0+2k\pi \Rightarrow \text{arg}(w) = \dfrac{k\pi}{1008}$$

Hence $w = e^{k\pi i/1008}$

And then $$w = \dfrac{z+i}{z-i} \Rightarrow z = \dfrac{-i(1+w)}{1-w}$$

We know that $w = e^{k\pi i/1008} = \cos \dfrac{k\pi}{1008} + i\sin \dfrac{k\pi}{2016}$

Denote $\dfrac{k\pi}{1008} = a$, now we expand and we have $$z = \dfrac{-i(1+\cos a + i\sin b)}{1- \cos a - i\sin a} = \dfrac{-i-i\cos a + \sin a}{1- \cos a - i\sin a}$$

Now recall we can use the denominator of $z$ and times it by the denominator's conjugate $$z = \dfrac{(-i-i\cos a + \sin a)(1-\cos a + i\sin a)}{(1- \cos a - i\sin a)(1-\cos a + i\sin a)} = \dfrac{\sin a}{1- \cos a} \in \mathbb{R}$$

I have came up with the above solution and i think it should be right, but then i read the answer and it says that $z$ is real if and only if $z = \overline{z}$. Now i dont dispute that but they claim that $\overline{z} = \dfrac{(-i)(-1-\overline{b})}{1-\overline{b}}$ where $b$ is just $e^{k\pi i/1008}$. I am confused over one thing, which is how they managed to know what $\overline{z}$ is. For me my understanding of conjugate of $z$ is changing the sign of $z$'s imaginary part. Which i clearly cant see any short cut?

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You have $$ \left|\frac{z+i}{z-i}\right|=1 $$ so $$ \frac{z+i}{z-i}=e^{2i\varphi} $$ for some $i$ (the role of $2\varphi$ will be clear later on). Solve for $z$: $$ z+i=ze^{2i\varphi}-ie^{2i\varphi} $$ and so $$ z=i\frac{e^{2i\varphi}+1}{e^{2i\varphi}-1}=i\frac{e^{i\varphi}+e^{-i\varphi}}{e^{i\varphi}-e^{-i\varphi}}= i\frac{2\cos\varphi}{2i\sin\varphi}=\tan\varphi $$ Thus you see that $2016$ was just a red herring.

You can also see it directly by writing $|z+i|=|z-i|$ as $$ (z+i)(\bar{z}-i)=(z-i)(\bar{z}+i) $$ and simplifying: indeed, $$ z\bar{z}+i\bar{z}-iz+1=z\bar{z}-i\bar{z}+iz+1 $$ becomes $$ z=\bar{z} $$

About your concerns: when you have written $$ z = \frac{-i-i\cos a + \sin a}{1- \cos a - i\sin a} $$ you have $$ \bar{z}=\frac{i+i\cos a+\sin a}{1-\cos a+i\sin a} $$ and you can check equality by cross multiplying. Your method is indeed better.

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    That is another great method of yours! thanks – 2017-02-04
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    Since $|(\dfrac{z+i}{z-i})^{2016}| = 1$, we also have $|\dfrac{z+i}{z-i}| = 1$. Thus $\dfrac{|z+i|}{|z-i|} = 1 \Rightarrow |z+i| = |z-i| \Rightarrow |z+i|^{2} = |z-i|^{2}$. Now expand and you get $(z+i)(\overline{z+i}) = (z-i)(\overline{z-i}) \Rightarrow... i(z-\overline{z}) = 0 \Rightarrow z = \overline{z}$ – 2017-02-04
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    @ilovewt Yes, it's the “direct method” I showed in my answer. Probably the easiest one, but the first method uses a trick that can turn out to be useful in several situations: when you have $e^{2i\varphi}+1$, you can write it as $e^{i\varphi}(e^{i\varphi}+e^{-i\varphi})=2e^{i\varphi}\cos\varphi$. Similarly for $e^{2i\varphi}-1$. – 2017-02-04
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This follows from Complex conjugate properties and that conjugate of $i$ is $-i$ $$z=\frac{-i(1+w)}{1-w}\\\overline{z}=\overline{\frac{-i(1+w)}{1-w}}=\frac{\overline{-i(1+w)}}{\overline{1-w}}=\frac{\overline{(-i)}\cdot\overline{(1+w)}}{1-\overline{w}}=\frac{i(1+\overline{w})}{1-\overline{w}}\\\overline{z}=\frac{-i(-1-\overline{w})}{1-\overline{w}}$$

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    Oh i finally see where i did not understand, i forgot the property that $\overline{(\dfrac{z_1}{z_2})}= \dfrac{\overline{z_1}}{\overline{z_2}} $ thanks! – 2017-02-04