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I need to solve $$iz^4\overline{z}+1=0$$ and write the result in algebric form.

To solve I tried to use the goniometric form:

$$i=\{cos\frac{\pi}{2}+ isin\frac{\pi}{2}\}$$ $$z^4 = \rho^4\{cos(4\theta)+isin(4\theta)\}$$ $$\overline{z} = \rho\{cos(-\theta)+isin(-\theta)\}$$ so I get:

$$\rho^5\{cos(3\theta+\frac{\pi}{2})+isin(3\theta+\frac{\pi}{2})\} + 1 = 0$$

$$\begin{cases} \rho^5 + 1 = 0 \\ 3\theta+\frac{\pi}{2} = 0\end{cases}$$ $$\begin{cases} \rho = -1 \\ \theta = -\frac{\pi}{6}\end{cases}$$

is this right? Does it means that all the solution are coincident? How should I write this in algebric form?

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Aren't you assuming $\rho\ge0$? You can get a more economical solution by using exponential notation $z=\rho e^{i\theta}$. You are aiming to solve $z^4\overline z=i$. This is $\rho^5e^{3i\theta}=e^{i\pi/2}$. Taking absolute values, $\rho^5=1$, so $\rho=1$. Then $3\theta=\pi/2+2k\pi$ (with $k\in\Bbb Z$) which gives you three values of theta between $0$ and $2\pi$. Of course, $\pi/6$ is one; what are the other two?

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Hint (algebraic, non-polar):  write the equation as $\;iz^4\overline{z}=-1\,$, then taking the magnitudes of the two sides gives $\,|z|^5 = 1 \implies |z|=1\,$, so $\,\bar z = 1/z\,$ and the equation reduces to:

$$i z^3+1=0 \;\;\iff\;\; z^3-i=0$$

To simplify the calculations, make the substitution $z = iw\,$, then the equation becomes:

$$-i w^3 - i =0 \;\;\iff\;\; w^3+1=0 \;\;\iff(w+1)(w^2-w+1)=0\;\;$$

Solve the latter for $w\,$, then substitute back into $z = iw$ for the roots in $z\,$.

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