TITLE: How to prove : If a complex number is equal to its own complex conjugate, then it's real.
QUESTION [1 upvotes]: \begin{align}
a+ib&=a-ib \\
a+i2b&=a \\
i2b&=0 \end{align}
To me, this seems weird. Any suggestions?

REPLY [5 votes]: Let $z=a+bi$ be a complex number, where $a,b \in \mathbb R$.
Then, it's complex conjugate is : $\bar{z}=a-bi$.
If $z=\bar{z}$ then : 
$$a+bi = a-bi \Leftrightarrow 2bi = 0 \Rightarrow bi = 0$$
Obviously, the imaginary unit $i$ is $\neq 0$, so it must be $b=0$.
But note that $b$ is the imaginary part of the complex number $z$ and more specifically : 
$$b = \Im\{z\} = 0$$
Since the imaginary part is zero, this means that $z$ is a real number.

REPLY [1 votes]: You're basically done. If $2bi = 0$, then $b=0$. Your number is of the form $a+bi$, so if $b=0$, the number is real (no imaginary part).

REPLY [0 votes]: $$
a+bi=a-bi\implies bi=-bi\implies i(bi)=-i(bi)\implies -b=b
$$
either $b$ is zero or $1=-1$ which is silly. So $b=0$.
You may also multiply 
$$
2bi=0
$$
on both sides by $\frac{-i}{2}$ to conclude $b=0$.
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