Solve $$\begin{cases}z_1z_2=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\\frac{z_1}{\overline{z_2}^2}=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}$$ over $\mathbb C$. The answer should be in trigonometric form. $z=a+ib$.

Let: $$ z_1=r_1(\cos\theta_1+i\sin\theta_1)\\ z_2=r_2(\cos\theta_2+i\sin\theta_2)\\ \frac{z_1}{\overline{z_2}^2}=\frac{z_1z_2^2}{\overline{z_2}^2z_2^2}=\frac{r_1r_2^2}{|z_2|^4}\stackrel{(\ast)}{=}\frac{r_1r_2^2}{r_2^4}=\frac{r_1}{r_2^2} $$ then: $$ \begin{cases}r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\[10pt] \dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2+i\sin2\theta_2)}\stackrel{(\ast\ast)}{=}\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases} $$

According to De Moivre's rule we know that $z^n=r^n \operatorname{cis}(n\theta)$ but in the transition $(\ast)$ why does $|z_2|^4=r_2^4$?

Lastly in the transition $(\ast\ast)$ is there some trig identity which leads to that?

Please explain the points $(\ast)$ and $(\ast\ast)$.

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    $\begingroup$ First, $\newcommand{\cis}{\text{cis}}|z^n|=|r^n\cis(n\theta)|=r^n|\cis(n\theta)|=r^n$. Second, $\dfrac{\cis(\theta)}{\cis(\psi)}=\cis(\theta-\psi)$, i.e it should be $\cos(\theta_1-2\theta_2)+i\sin(\theta_1-2\theta_2)$ and not as written. $\endgroup$ – Galc127 Jul 7 '17 at 7:43
  • $\begingroup$ @Galc127 in $|z^n|=r^n|cis(n\theta)|$ what happens with $|cis(n\theta)|$? Also confirming that it's $cis(\theta_1+2\theta_2)$ $\endgroup$ – Yos Jul 7 '17 at 7:48
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    $\begingroup$ 1. $\newcommand{\cis}{\text{cis}}|\cis(n\theta)|=|\cos(n\theta)+i\sin(n\theta)|=\sqrt{\cos^2(n\theta)+\sin^2(n\theta)}=1$. 2. It is $\cis(\theta_1+2\theta_2)$ because you have $$\frac{z_1}{\bar{z_2}^2}=\frac{r_1\cis(\theta_1)}{r_2^2\cis(-2\theta_2)}=\frac{r_1}{r_2^2}\cis(\theta_1+2\theta_2)$$ The solution missed the fact that it is $\bar{z_2}$ in denominator... $\endgroup$ – Galc127 Jul 7 '17 at 7:55
  • $\begingroup$ The general relation you want is $(a ~\cis~ b) \times (c ~\cis~ d) = ( (ab) ~\cis~ (c + d))$. This isn't a consequence of demoivre's , but rather demoivre is a special case of the above relation, which can be proven with straightforward angle sum formulas. Demoivre is actually pretty irrelevant to this problem. $\endgroup$ – DanielV Jul 7 '17 at 8:10

$r_1r_2=10$ and $\frac{r_1}{r_2^2}=\frac{2}{25}$, which gives $r_1=2$ and $r_2=5$.

Let $z_1=2cis\theta_1$ and $z_2=5cis\theta_2$.

Thus, $\theta_1+\theta_2=\frac{4\pi}{5}+2\pi k$ and $\theta_1+2\theta_2=\frac{\pi}{5}+2\pi m$, where $\{k,m\}\subset\mathbb Z$.

Finally we obtain: $\theta_1=\theta_2=\frac{7\pi}{5}$

Your $*$ is true by the De Moivre's rule.

Your $**$ is true because $\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ and $\overline{cis\theta}=cis(-\theta)$.

$\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ because $$\frac{cis\theta}{cis\phi}=\frac{cis\theta\overline{cis\phi}}{cis\phi\overline{cis\phi}}=cis(\theta-\phi)=$$

  • $\begingroup$ Please read the last line in the OP. I want to understand the points $(\ast)$ and $(\ast\ast)$, I know the final answer. $\endgroup$ – Yos Jul 7 '17 at 7:45
  • $\begingroup$ @Yos I fixed. See now. $\endgroup$ – Michael Rozenberg Jul 7 '17 at 7:56
  • $\begingroup$ Why is $\frac{cis\theta_1}{cis\theta_2}=cis(\theta_1-\theta_2)$? Is there an identity for that? $\endgroup$ – Yos Jul 7 '17 at 8:01
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    $\begingroup$ @Yos See now please. $\endgroup$ – Michael Rozenberg Jul 7 '17 at 8:18
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    $\begingroup$ @Yos Yes, but I think it's better to use the rule: $\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ $\endgroup$ – Michael Rozenberg Jul 7 '17 at 8:28

The line where you have $(*)$ should better be $$ \left|\frac{z_1}{\overline{z_2}^{\,2}}\right|= \frac{|z_1|}{\bigl|\overline{z_2}^{\,2}\bigr|}=\frac{r_1}{r_2^2} $$ There is no reason for the equality before the one you mark with $(*)$.

The equality you mark with $({*}{*})$ is indeed wrong and it should be $$ \frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2-i\sin2\theta_2)}=\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}) $$ because if $z=r(\cos\theta+i\sin\theta)$, then $\bar{z}=r(\cos\theta-i\sin\theta)$. Next apply the standard rules \begin{gather} \cos\theta-i\sin\theta=(\cos\theta+i\sin\theta)^{-1}\\[4px] (\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)= \cos(\alpha+\beta)+i\sin(\alpha+\beta) \end{gather}

Further notes.

The equality $$ \frac{z_1z_2^2}{\overline{z_2}^{\,2}z_2^2}=\frac{r_1r_2^2}{|z_2|^4} $$ is generally false, because there's no reason for $z_1z_2^2$ to be real.

When you expand $\dfrac{z_1}{\overline{z_2}^{\,2}}$, you should write $$ \frac{r_1(\cos\theta_1+i\sin\theta_1)} {\bigl(\overline{r_2(\cos\theta_2+i\sin\theta_2}\bigr)^2} = \frac{r_1(\cos\theta_1+i\sin\theta_1)} {(r_2(\cos\theta_2-i\sin\theta_2)^2} = \frac{r_1(\cos\theta_1+i\sin\theta_1)} {r_2^2(\cos2\theta_2-i\sin2\theta_2)} $$

  • $\begingroup$ Can you please explain why $\ast\ast$ equality is a mistake? As far as I understand we have $$\frac{\cos\theta_1+i\sin\theta_1}{\cos 2\theta_2+i\sin 2\theta_2}=\frac{(\cos\theta_1+i\sin\theta_1)(\cos 2\theta_2-i\sin 2\theta_2)}{(\cos 2\theta_2+i\sin 2\theta_2)(\cos 2\theta_2-i\sin 2\theta_2)}=...$$ $\endgroup$ – Yos Jul 7 '17 at 8:36
  • $\begingroup$ @Yos In the denominator there is $\overline{z_2}^{\,2}$, not $z_2^2$. $\endgroup$ – egreg Jul 7 '17 at 8:37
  • $\begingroup$ yes but I'm referring to the result of this: $$\frac{z_1}{\overline{z_2}^2}=\frac{z_1z_2^2}{\overline{z_2}^2z_2^2}=\frac{r_1r_2^2}{|z_2|^4}\stackrel{(\ast)}{=}\frac{r_1r_2^2}{r_2^4}=\frac{r_1}{r_2^2}\Rightarrow \frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos 2\theta_2+i\sin 2\theta_2)}$$ $\endgroup$ – Yos Jul 7 '17 at 8:38
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    $\begingroup$ @Yos If you start from a wrong equality, you get nothing good, do you? It is false that $z_1z_2^2=r_1r_2^2$. The “solution” has two gross errors; the final result is correct, but just by chance. $\endgroup$ – egreg Jul 7 '17 at 8:50
  • $\begingroup$ Thanks for the additional note in the answer! $\endgroup$ – Yos Jul 7 '17 at 9:34

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