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


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?


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?

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

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


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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.