Proving that a complex number $z$ is real.

Question

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?

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

score 4 · Answer 1 · answered Mar 15 '13 at 10:41

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.

score 2 · Accepted Answer · answered Mar 15 '13 at 10:43

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, $z=x$.

score 2 · Answer 3 · answered Mar 15 '13 at 10:59

2

$0$ 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.

answered Mar 15 '13 at 10:59

David Alspaugh

21

score 0 · Answer 4 · edited Mar 30 '17 at 17:45

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:)

Proving that a complex number $z$ is real.

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