# Complex numbers and conjugates. [closed]

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 LinJan 12 at 9:49

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

HINT

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 $$\bar z z=|z|^2$$.

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

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

$$\sqrt{15}=2\sqrt3e^{-it}+\dfrac3{i\sqrt3e^{it}}=\sqrt3(2-i)e^{-it}$$

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

We are done.

We can go even further.

$$e^{it}=e^{-i\arcsin\dfrac1{\sqrt5}}$$

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