Proving that a complex number $z$ is real. A problem I have in my book is to prove that $z$ is real if and only if $\bar{z} = z$.
So far I have got that for $z = x + iy$, if $z$ is real, $y = 0$ and thus $z = x = \bar{z}$ as  $\bar{z} = x - iy$ where $y = 0$ (if I'm right).
Now my book mentions something like converse of this, i.e, if $\bar{z} = z$ then $x+iy = x-iy$, where the last equality implies $y = -y$ and thus $y = 0$ (I don't get what equality it's talking about).
Also later it's explained that, therefore, $z = x$ and thus is real. (I don't get the second part at all).
Can someone please help me to understand this?
 A: Notice that when we are saying two complex number are equal, we mean that both the real parts and imaginary parts of the two numbers are equal. So $$z=\bar{z}\iff x+iy=x-iy\iff\left\{
\begin{array}{ccl}x=x\\y=-y\end{array}\right.\iff y =0\iff z=x$$
So $z=\bar{z}$ implies $z$ is real. 
A: $0$ is the only number in the reals which is equal to its own negative. The proof showed both equalities because it is proving an if and only if statement.
A: let $z=x+iy$  where, $x,y∈\mathbb{R}$
First assume that we are given $z$ is real then its imaginary part must be equal to $0$
Therefore, $z=x$.
Now, assume that $z=\overline{z}$ where  $\overline{z}=x-iy$
then, $$x+iy=x-iy$$ $$iy=-iy$$ $$2y=0$$  that means $y=o$ .
So,we get, $z=x$.   
A: Just a thought. what happen if $z=\cos(i^2)$;
then, 
derivative of $\cos(x^2)$;
then, Taylor series
compile it up into sigma form,
then,
maybe 
a variable becomes a real, as a sequence, with endless fluctuation as a function of $\cos$;
then,
maybe, if lucky
become a irrational number, I mean, at least, it is real number;
e limit
compile it up into a symble, 
You got your own symble :)
Best luck:)
