# For given equation, calculate modulus of complex number $z$

If $z$ is a complex number such that Real part of $z\neq 2$ and

$$z^2=4z+|z|^2+\frac{16}{|z|^3}$$

I assumed $z=x+iy$ and tried equation real and imaginary part on both sides.

After equating imaginary part, I got $x=2$ or $y=0$ but when I equal real parts, I am getting ugly calculations. Could someone suggest a better approach?

## 2 Answers

You have that $z^2-4z$ is real, so $z^2-4z=\bar{z}^2-4\bar{z}$ and therefore $$(z+\bar{z})(z-\bar{z})=4(z-\bar{z})$$ which implies $z=\bar{z}$ or $z+\bar{z}=4$.

In the first case, $z$ is real, so the equation simplifies greatly. In the second case, the real part of $z$ would be $2$: can you see it?

• Yes I can. This approach is so much better as compared to mine.. – Mathgeek Jan 16 '17 at 19:30

If you take $z=a+ib$ and substitute

$$a^2-b^2+2abi=4a+4bi+a^2+b^2+\frac{16}{(a^2+b^2)\sqrt{a^2+b^2}}$$

The imaginary part must be equal, so $2ab=4b$. The restriction of $a\neq 2$ determine that $b=0$. The last equation is reduced to

$$a^2=4a+a^2+\frac{16}{a^3} \Rightarrow a+\frac{4}{a^3}= 0$$

Assuming that $a\neq 0$ (because, in this case, $z=0$ and the right part of the equation of the exercise would be unbounded), you get

$$-4=a^4$$

This is a contradiction, because $a\in \mathbb{R}$, so there are no numbers which solve this equation with this restriction.