# Equation in the complex plane $8z=i|z|^3\bar{z}$?

I do not know what to do with this equation. I tried to make both sides in trigonometric form, but after I don't know how to move forward to solve it. $$8z=i|z|^3\bar{z}$$ In trigonometric form $$8\varphi[\cos(\theta) + i \sin(\theta)]=\varphi^4[\cos(\frac{\pi}{2}-\theta) + i \sin(\frac{\pi}{2}-\theta)]$$ Now I don't know how to continue.

• $z=0$ seems to be a obvious solution. – Fabian Jan 11 '16 at 10:51
• If one takes the norm of both sides, one gets $8 |z| = |z|^4$, so either $|z| = 0$, in which case $z = 0$ (as Fabian notes, it's easy to check that this is a solution) or $|z| = 2$. Now, write a generic complex number with modulus $2$ as $z = 2 e^{i \theta}$, substitute, and solve. – Travis Jan 11 '16 at 10:54
• Another option: Let $z=re^{i\theta}$, thus the equation is $8re^{i\theta}=ir^3\cdot{re^{-i\theta}}=ir^4e^{-i\theta} \Rightarrow r\left[8e^{i\theta}-ir^3e^{-i\theta}\right]=0$. Now you have two options. Use the fact that $-i=e^{\frac{3}{2}\pi i}$ and solve. – Galc127 Jan 11 '16 at 10:55
• How about $z=2_{\pi/4}$?. I have solved this by saying $z = r_\theta$. – JnxF Jan 11 '16 at 10:57

Let $r=|z|$ and $\theta=\arg(z)$ so that $z=re^{i\theta}$. Then $\overline{z}=re^{-i\theta}$ and so the equation becomes $$8re^{i\theta}=ir^4e^{-i\theta}.$$ Note that $i=e^{i\tfrac{\pi}{2}}$, so the right-hand side equals $r^4e^{i(\tfrac{\pi}{2}-\theta)}$. Rearranging terms the above becomes $$re^{i\theta}\cdot\left(8-r^3e^{i(\tfrac{\pi}{2}-2\theta)}\right)=0,$$ which shows that either $r=0$ or $$8-r^3e^{i(\tfrac{\pi}{2}-2\theta)}=0,$$ in which case $r=2$ and $\tfrac{\pi}{2}-2\theta=2k\pi$ for some $k\in\Bbb{Z}$. Then $\theta=(\tfrac14-k)\pi$, so $\theta=\tfrac14\pi$ or $\theta=\tfrac54\pi$ as $\theta\in[0,2\pi)$. This shows that there are precisely three solutions, which are $$z=\pm\sqrt{2}(1+i)\qquad\text{ or }\qquad z=0.$$
We see that $z=0$ is a trivial solution. Let $z=re^{i\theta+i2\pi k }$. Then $$8re^{i\theta+i2\pi k}=ir^4e^{-i\theta-i2\pi k},$$
giving $$8e^{i2\theta+i4\pi k}=r^3e^{i\pi/2+i2\pi\ell}.$$ Hence $r=2$ and $2\theta+4\pi k=\pi/2+2\pi\ell\implies\theta=\pi/4+\pi\ell-2\pi k\equiv \pi/4+\pi\ell$.
So we have $z=2e^{i(\pi/4+\pi\ell)}$, where $\ell\in\mathbb{Z}$. This gives the solutions $$\left\{0,2 e^{\frac{i \pi }{4}},2 e^{-\frac{i3}{4}\pi}\right\}.$$