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Let z = a + bi, where a and b are real numbers. If z/z* = c + di, where c and d are real, prove that c^2 + d^2 = 1.

I'm stuck and keep going round in circles. Any help would be much appreciated. Thanks.

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    $\begingroup$ Do you know about the modulus of a complex number? And do you know that $|a/b| = |a| / |b|$? $\endgroup$ – John Hughes Mar 21 '17 at 17:25
  • $\begingroup$ Can you prove that $z$ and $z^*$ have the same magnitude? Then the ratio will have magnitude 1. $\endgroup$ – NickD Mar 21 '17 at 17:26
  • $\begingroup$ hold on ignore comment I was thinking something else xD $\endgroup$ – Dan90 Mar 21 '17 at 17:34
  • $\begingroup$ How do you get $c^2+d^2$ if $c+di=(a^2 + 2abi - b^2)/(a^2 + b^2)$? $\endgroup$ – Sentinel135 Mar 21 '17 at 17:43
  • $\begingroup$ Multiply by c-di $\endgroup$ – m.bazza Mar 21 '17 at 17:46
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$\frac{z}{z*}= \frac{a+bi}{a-bi}$ which is the same as $\frac{a+bi}{a-bi} \frac{a+bi}{a+bi} = \frac{(a+b i)^2 }{(a+bi)(a-bi)}= \frac{a^2+2ab i-b^2}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}+i \frac{2ab}{a^2+b^2}$ which means that $c=\frac{a^2-b^2}{a^2+b^2}$ and $d=\frac{2ab}{a^2+b^2}$ so $c^2+d^2 = \frac{(a^2-b^2)^2+(2ab)^2}{(a^2+b^2)^2} = \frac{a^4-2a^2b^2+b^4+4a^2b^2}{(a^2+b^2)^2} = \frac{(a^2+b^2)^2}{(a^2+b^2)^2} = 1$

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Much shorter solution following the idea of John Hughes ($z\ne 0$ is required): $$|z^*| = |z|\implies |z/z^*| = |z|/|z^*| = 1.$$

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Let me show you a different way than Ahmad. Though his way works just as well. Let $z= a+bi$ where $a,b\in \mathbb R$. Then, $$\frac {z}{z^*}= \frac {a+bi}{a-bi}= \frac{a+bi}{a-bi}\cdot \frac{a+bi}{a+bi}= \frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i=c+di .$$ So for $c^2+d^2 = (c+di)(c-di)$ we have $$\frac{(a^2-b^2)+2abi}{a^2+b^2}\cdot\frac{(a^2-b^2)-2abi}{a^2+b^2}= \frac{a+bi}{a-bi}\cdot \frac{(a-bi)^2}{(a+bi)(a-bi)}=1$$

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