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Problem: I would like to solve the equation $$\overline{z}^4+z^2=16i$$ It seems really easy to solve but... I try the cartesian form, the polar form and the exponential form but nothing happens. I try to moltiplicate both sides for $z^4$ so $z^4\overline{z}^4=(z\overline{z})^4=|z|^8$ but this doesn't help. I try the substitution $z^2=t$ too... Hopeless. Wolfram alpha gives the solutions with approximation.

Any help will be appreciate.

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    $\begingroup$ Let $z^2=re^{it}$ or $a+ib$ $\endgroup$ – lab bhattacharjee Nov 6 '17 at 14:05
  • $\begingroup$ @labbhattacharjee I try it both, but the system is not easy to solve. $\endgroup$ – Ixion Nov 6 '17 at 14:08
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Let $z^2=x+iy$. The equation reads

$$\begin{cases}x^2-y^2+x=0,\\-2xy+y=16.\end{cases}$$

From the first, $y^2=x^2+x$ and plugging in the second squared,

$$(1-2x)^2(x^2+x)=256,$$

$$4x^4-3x^2+x-256=0.$$

This quartic equation can be solved analytically, but doesn't seem to lead to a nice solution. It has two distinct real roots, one of which yields a negative value of $x^2+x$. Hence, there is just one solution to the $z^2$ equation, and two of them after taking the square root.

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Let $z=x+yi$, where $\{x,y\}\subset\mathbb R$.

Thus, $$(x-yi)^4+(x+yi)^2=16i$$ or $$x^4-4x^3yi-6x^2y^2+4xy^3i+y^4+x^2-y^2+2xyi=16i,$$ which gives $$-2xy(x^2-y^2)+xy=8$$ and $$x^4-6x^2y^2+y^4+x^2-y^2=0.$$ Now, $$x^2-y^2=-\frac{8-xy}{2xy},$$ which gives $$\frac{(8-xy)^2}{4x^2y^2}-4x^2y^2-\frac{8-xy}{2xy}=0$$ and we get a very ugly numbers.

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  • $\begingroup$ Very ugly numbers, you're right. Well, now I think that the problem is the equation itself, maybe a typo... $\endgroup$ – Ixion Nov 6 '17 at 14:31

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