Find all the solutions to the equation:

My attempt:
$z^3\bar{z}=3+4i \rightarrow arg(z^3\bar{z})=arg(3+4i)\\$
$arg(z^3)-arg(z)=\tan^{-1}(4/3)+2\pi k\\$
$2arg(z)=\tan^{-1}(4/3)+2\pi k\\$
Therefore the angle is:
$arg(z)=\frac{\tan^{-1}(4/3)}{2}+\pi k$

And to find the magnitude:
$| z^3\bar{z}|=|z^3||z|=|z|^4=|3+4i| \\$
$|z|^4=5 \rightarrow |z|=\sqrt[4]5\\$
$z=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)+i\pi k}.$
By plugging $k = 0$ and $k = 1$ we get
$z_{1}=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)}=\sqrt[4]5(\cos(0.5\tan^{-1}(4/3)) + i\sin(0.5\tan^{-1}(4/3))$ $z_{2}=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)+i\pi}\\=\sqrt[4]5(\cos(0.5\tan^{-1}(4/3)+\pi) + i\sin(0.5\tan^{-1}(4/3)+\pi)$

Yet the correct solution is:
$z=\pm \sqrt[4]5(\cos(0.5\tan^{-1}(4/3)) + i\sin(0.5\tan^{-1}(4/3))\\$
So $z_1$ is a correct solution, but i'm not sure what about $z_2$ and why there is a negative solution in the answer.


Simplify $z_2$ using $cos(\theta+\pi)=-cos(\theta)$ and $sin(\theta+\pi)=-sin(\theta)$.


You can simplify $z^3\bar z=\sqrt{5}z^2$ giving $$z^2=\dfrac{3+4i}{\sqrt{5}}\iff z=\pm\dfrac{2+i}{\sqrt[4]{5}}$$

Since $(2+i)^2=4+4i-1=3+4i$

  • $\begingroup$ Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any? $\endgroup$ – user376343 Jan 20 at 23:36

From$$r^3r\text{ cis }(3\theta-\theta)=5\text{ cis}\arctan\frac43,$$ we draw $$r=\sqrt[4]5,$$



$$\pm\sqrt[4]5\text{ cis}\left(\frac12\arctan\frac43\right).$$

(The $\pm$ comes from $k\pi$.)


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