# Proving that a complex number is nonreal.

Let $m$ be a nonzero complex number such that and $z=-1+im$ and $w=-1-im$. Prove that the number $$\frac{m-w}{z-w}$$ is nonreal. I've tried all sorts of approaches to this question but there seems to be something I'm missing. Any help would be appreciated, thank you in advance. EDIT: My bad, I made a mistake in the text.

• I think you need to do the basic arithmetics and then show that your number $a = \frac{m-w}{z-w}$ can be written in the following form $a = x + iy$ Feb 27, 2018 at 10:52
• For $m=-i$ this seems to be false Feb 27, 2018 at 10:52

This statement seems to be wrong.

$${m-\omega\over z-\omega}={-i\over 2m}+{-(1+i)i\over 2}={-i\over 2m}+{1-i\over 2}.$$

Now, write $m$ as $m=x+iy\;$ for some real numbers $x$ and $y$ with $x\neq 0\neq y$. Then,

$${-i\over 2m}+{1-i\over 2}={1-i\over 2} - {i\over 2(x + i y)}.$$

Out of this you get that, $$\text{Re}\left({m-\omega\over z-\omega}\right)={1\over 2}-{y\over 2(x^2+y^2)},$$ which is equal to zero only for certain values of $x$ and $y$. Therefore is not true that ${m-\omega\over z-\omega}$ is imaginary.

• This was helpful even though I made a mistake in the text, thank you. Feb 27, 2018 at 11:38
• The real part of the expression you get isn't $\frac{1}{2}$, it is $\frac{1}{2} + \frac{y}{2(x^2 + y^2)}$. It is indeed $\frac{1}{2}$ iff $y=0$.
– Itai
Feb 27, 2018 at 14:50
• @sillyfly Yes, you're completely right. Edited the answer. Thanks.
– Edu
Feb 27, 2018 at 14:57

$$\forall m \in \mathbb{R}^*, Re\left(\frac{m-w}{z-w}\right)=\frac{1}{2}\ne 0$$ either you made a typo or the claim is wrong