Help me please, I can't resolve it. I don't have any idea.
If $z,w \in {\mathbb C}\setminus\{0\}$, prove that $zw$ is real if and only if $\exists{k}\in {\mathbb R}$ such that $w=k\bar{z}$.
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At least, show what you've figured out so far and where you have stuck exactly. – Ehsan M. Kermani Oct 08 '13 at 21:52
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And note that on the right hand side, there is only a single candidate for $k$, namely $w / {\bar z}$. – Magdiragdag Oct 08 '13 at 21:55
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Hint: $z$ is real iff $\Im z = 0$ Now show 1. $$\Im (z (k\bar z)) = 0 \quad \forall\ k\in\mathbb R$$ and 2. $$\Im (zw) = 0 \Rightarrow w = k\bar z$$ – AlexR Oct 08 '13 at 21:56
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Let's write $z=a+bi$ and $w=c+di$ so that $zw=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. From this, it's clear that $zw$ is real if and only if $ad+bc=0$
Now there are two cases to check. Suppose $a\neq 0$ and so $ad+bc=0$ if and only if $-\frac{c}{a}b=d$. Let $k=\frac{c}{a}$ and so $-\frac{c}{a}b=d$ if and only if there is a $k\in\mathbb{R}$ such that $ka=c$ and $-kb=d$. This is equivalent to saying there exists a $k$ such that $ka-kbi=c+di$ which happens if and only if there exists a $k$ such that $w=k\bar{z}$.
The second case is if $a=0$. In this case $ad+bc=0$ if and only if $bc=0$ and so either $b=0$ or $c=0$. But $b\neq 0$ because $z\in\mathbb{C}\setminus\{0\}$ and so $bc=0$ if and only if $c=0$. This occurs if and only if both $z=bi$ and $w=di$ with $b\neq0\neq d$ which happens if and only if there exists a real $k$ such that $k=-\frac{d}{b}$ if and only if there eixsts a $k$ such that $-kbi=di$ if and only if there exists a $k$ such that $w=k\bar{z}$.
Dan Rust
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Hints. What do you know about $z\bar{z}$?
(1) If $zw=a$ is real, then $z\bar{z}w=a\bar{z}$ and so …
(2) If $w=k\bar{z}$ with $k$ real, then $zw=kz\bar{z}$ and so …
egreg
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$zw \in {\mathbb R} \iff (w / \bar z) z \bar z \in {\mathbb R} \iff w / \bar z \in {\mathbb R}$; the second equivalence holds because $z {\bar z} = |z|^2 \in {\mathbb R}$.
Magdiragdag
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