Let $z \in \mathbb{C}$, show that if $\left(\dfrac{z+i}{z-i}\right)^{2016}=1$ then $z$ is a real number. 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?
 A: 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.
A: This follows from Complex conjugate properties and that conjugate of $i$ is $-i$ 
$$z=\frac{-i(1+w)}{1-w}\\\overline{z}=\overline{\frac{-i(1+w)}{1-w}}=\frac{\overline{-i(1+w)}}{\overline{1-w}}=\frac{\overline{(-i)}\cdot\overline{(1+w)}}{1-\overline{w}}=\frac{i(1+\overline{w})}{1-\overline{w}}\\\overline{z}=\frac{-i(-1-\overline{w})}{1-\overline{w}}$$
