Given that $|z|=√3$, solve the equation $$2\overline{z}+\frac3{iz}=\sqrt{15}.$$

How to solve this question without a calculator?


closed as off-topic by Saad, Abcd, Cesareo, metamorphy, Lee David Chung Lin Jan 12 at 9:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Cesareo, metamorphy, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be. $\endgroup$ – Aleksa Nov 11 '18 at 11:16
  • $\begingroup$ @Vittal Kamath, so what is the answer did you get? $\endgroup$ – Dhamnekar Winod Nov 11 '18 at 12:12


Multiplying by $z$ we obtain

$$2\bar z+\frac3{iz}=\sqrt{15} \implies 2\bar zz+\frac3{iz}z\frac i i=\sqrt{15}z$$

then recall that $\bar z z=|z|^2$.

  • $\begingroup$ ,what are the next steps to arrive at final answer? $\endgroup$ – Dhamnekar Winod Nov 11 '18 at 12:13
  • $\begingroup$ @DhamnekarWinod What did you obtain from here $2\bar zz+\frac3{iz}z\frac i i=\sqrt{15}z$? $\endgroup$ – gimusi Nov 11 '18 at 12:14
  • $\begingroup$ ,I got $6+ \frac{3}{i}=\sqrt{45}$ $\endgroup$ – Dhamnekar Winod Nov 11 '18 at 12:16
  • $\begingroup$ @DhamnekarWinod Why that? We have $\bar z z=|z|^2$ and here we are ok, then $\frac 3 i =-3i$ but at the RHS we should have $\sqrt{15}z$. $\endgroup$ – gimusi Nov 11 '18 at 12:21
  • $\begingroup$ because|z|=$\sqrt{3}$. If this is wrong,then $z=\frac{6-3i}{\sqrt{15}}$ $\endgroup$ – Dhamnekar Winod Nov 11 '18 at 12:31

WLOG $z=\sqrt3e^{it}\implies\bar z=\sqrt3e^{-it}$ where $t$ is real


$$\iff e^{it}=\dfrac{2-i}{\sqrt5}$$

We are done.

We can go even further.


$$\implies t=2n\pi -\arcsin\dfrac1{\sqrt5}$$ where $n$ is any integer


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