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I need help with two equation involving complex numbers

$(1)$ $\frac{z+1}{z-1}=i|z|$

I tried to solve it in this way

Supposing $z\neq1$

$z+1=i|z|(z-1)$

$z=\frac{-1-|z|}{1-|z|}$

But I don't know how to continue

$(2)$ $6z^3+5|z|^2=6(Im(z))^2$

In this one I tried with the substitution $z=a+ib$ but calculation are not easy so I would like to avoid it and solve it in another way (if there is one)

Thanks a lot in advice

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  • $\begingroup$ Did you try writing $z=a+bi$ and then translating $z+1=i|z|(z-1)$ into two equations in $a$ and $b$? $\endgroup$ – Gregory Grant Nov 22 '15 at 0:37
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Let $\phi(z) = {z+1 \over z-1}$, note that $\phi(\phi(z)) = z$.

To see what $z$ are mapped to the imaginary axis, consider $\phi(\lambda i) = - { 1 \over 1+ \lambda^2} (1-\lambda^2 + 2 i \lambda )$.

Hence the only $z$ that are mapped to the imaginary axis are $\pm i$. We have $\phi(i) = -i, \phi(-i) = i$, hence $i$ is the solution.

For the other problem, some computation solves the problem. Note that $6 (\operatorname{im} z)^2-5 |z|^2$ is real, hence $z^3$ must be real as well. Hence we can look for solutions of the form $x$, $x ({1 \over 2} \pm i { \sqrt{3} \over 2})$, where $x$ is real.

Let $f(z) = 6 z^3 -6 (\operatorname{im} z)^2+5 |z|^2$, then $f(x) = 6 x^3 + 5 x^2 = x^2(6x+5)$ which has solutions $0, -{5 \over 6}$. Similarly, but more tediously, solving $f(x({1 \over 2} + i { \sqrt{3} \over 2}) ) = 0$ gives solutions $0, {1 \over 12}$ and similarly for the other equation which gives $0, {1 \over 12}$.

Hence the solutions are $z \in \{ 0, -{5 \over 6}, {1 \over 12} ({1 \over 2} + i { \sqrt{3} \over 2}), {1 \over 12} ({1 \over 2} - i { \sqrt{3} \over 2})\}$.

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  • $\begingroup$ Thanks for the detailed answer, I don't understand why in the problem $2)$ you looked for solutions in the form $x(\frac{1}{2}±i\frac{\sqrt{3}}{2}}$ besides the ones in the form $x$ (which are clear to me). And why this particular form? Is there a $\frac{\pi}{3}$ angle that I don't see? $\endgroup$ – Gianolepo Nov 23 '15 at 8:31
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    $\begingroup$ Yes, since $z^3$ is real, then $z$ must have the form $r e^{i \theta}$ where $\theta = k { \pi \over 3}$. Look at the cube roots of $\pm 1$. $\endgroup$ – copper.hat Nov 23 '15 at 8:34
  • $\begingroup$ Ok! One last thing, in problem $1)$ I don't understand how the fact that $\phi(\phi(z))=z$ is related to the problem, I mean since $\frac{z+1}{z-1}=i|z|$ is it possible to conclude directly that $z$ will be in the form $z=\lambda i$ ("completely imaginary") ? $\endgroup$ – Gianolepo Nov 23 '15 at 13:43
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    $\begingroup$ You are trying to find $z$ such that $\phi(z)= \lambda i$ for some $\lambda$ (which depends on $z$, of course). So, we solve this equation for $\lambda$. The above equality shows that $z=\phi(\phi(z)) = \phi(\lambda i)$, and expanding this gives an equation in $\lambda$. Then we look for values of $\lambda$ that make $\phi(\lambda i)$ purely imaginary. $\endgroup$ – copper.hat Nov 23 '15 at 15:24
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    $\begingroup$ It does. The main requirement is that $z^3$ is real. I should have used $r$ rather than $x$ to be suggestive, but it simplifies life a little to allow $x$ to be negative as well ($x$ is real). Then instead of dealing with 6 cases (and a positivity constraint), I can deal with 3 cases. $\endgroup$ – copper.hat Nov 27 '15 at 14:59
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$$z=\frac{i\left| z \right|+1}{i\left| z \right|-1}\Rightarrow \left| z \right|=\frac{\left| i\left| z \right|+1 \right|}{\left| i\left| z \right|-1 \right|}=\frac{\sqrt{1+{{\left| z \right|}^{2}}}}{\sqrt{1+{{\left| z \right|}^{2}}}}=1$$ Return $$\frac{z-1}{z+1}=i$$

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A quick way to solve equation (1) would be to observe, that $z\mapsto\frac{1+z}{z-1}$ is bijective and maps the boundary of the unit circle onto the imaginary axis (and 1 to $\infty$). Then you can just drop the term $\vert z\vert$ on the right side to solve the equation.

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