Let be $a,b,c$ complex numbers such that $|a|=|b|=|c|=r$ then show that $\left | \frac{ab+bc+ca}{a+b+c}\right|=r$
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Tags: complex numbers
claserken
09.01.2017 18:51
We use the well known fact that $z\cdot\overline{z}=|z|^2$. We get that $\overline{a}=\frac{r^2}{a}$ and similarly for $b$ and $c$. It suffices to show that $\frac{ab+bc+ca}{a+b+c}\cdot \overline{\left (\frac{ab+bc+ca}{a+b+c}\right )}=r^2$. This is equivalent to $\frac{ab+bc+ca}{a+b+c}\cdot \frac{\frac{r^2}{ab}+\frac{r^2}{bc}+\frac{r^2}{ca}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{ab+bc+ca}{a+b+c}\cdot \frac{\frac{r^2}{ab}+\frac{r^2}{bc}+\frac{r^2}{ca}}{\frac{ab+ac+bc}{abc}}$ which easily simplifies to $\frac{ar^2+br^2+cr^2}{a+b+c}=r^2$, so we are done$\blacksquare$
User335559
05.08.2017 18:42
claserken wrote:
We use the well known fact that $z\cdot\overline{z}=|z|^2$. We get that $\overline{a}=\frac{r^2}{a}$ and similarly for $b$ and $c$. It suffices to show that $\frac{ab+bc+ca}{a+b+c}\cdot \overline{\left (\frac{ab+bc+ca}{a+b+c}\right )}=r^2$. This is equivalent to $\frac{ab+bc+ca}{a+b+c}\cdot \frac{\frac{r^2}{ab}+\frac{r^2}{bc}+\frac{r^2}{ca}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{ab+bc+ca}{a+b+c}\cdot \frac{\frac{r^2}{ab}+\frac{r^2}{bc}+\frac{r^2}{ca}}{\frac{ab+ac+bc}{abc}}$ which easily simplifies to $\frac{ar^2+br^2+cr^2}{a+b+c}=r^2$, so we are done$\blacksquare$
Hi, I'm bad on complex numbers, but why did you remove the overline from the 4th line to the 5th?
alexheinis
07.08.2017 16:57
By homogeneity you may assume $|a|=|b|=|c|=1$. Then $|ab+ac+bc|=|abc\overline{c}+abc\overline{b}+abc\overline{a}|=|\overline{a}+\overline{b}+\overline{c}|=|a+b+c|$.