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.
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$\begingroup$ 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 $a = x + iy$ $\endgroup$– HendrraCommented Feb 27, 2018 at 10:52
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1$\begingroup$ For $m=-i$ this seems to be false $\endgroup$– charmdCommented Feb 27, 2018 at 10:52
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2 Answers
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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.
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1$\begingroup$ This was helpful even though I made a mistake in the text, thank you. $\endgroup$ Commented Feb 27, 2018 at 11:38
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1$\begingroup$ 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 $y=0$. $\endgroup$– ItaiCommented Feb 27, 2018 at 14:50
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$\begingroup$ @sillyfly Yes, you're completely right. Edited the answer. Thanks. $\endgroup$– EduCommented Feb 27, 2018 at 14:57
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$$\forall m \in \mathbb{R}^*, Re\left(\frac{m-w}{z-w}\right)=\frac{1}{2}\ne 0$$ either you made a typo or the claim is wrong
