# 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).

• Notice that two complex numbers $a+ib$ and $x+iy$ are equal iff $a=x$ and $b=y$. Now what can you conclude from $x+iy = x-iy$ ? – Jean-Claude Arbaut Mar 15 '13 at 10:47

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, $z=x$.

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.

$0$ 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.

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:)