Given that $|z|=√3$, solve the equation $$2\overline{z}+\frac3{iz}=\sqrt{15}.$$
How to solve this question without a calculator?
Asked
Active
Viewed 149 times
1
user10354138
- 33,239
-
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? – Win_odd Dhamnekar Nov 11 '18 at 12:12
2 Answers
4
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 $\bar z z=|z|^2$.
user
- 154,566
-
-
@DhamnekarWinod What did you obtain from here $2\bar zz+\frac3{iz}z\frac i i=\sqrt{15}z$? – user Nov 11 '18 at 12:14
-
-
@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 $\sqrt{15}z$. – user Nov 11 '18 at 12:21
-
because|z|=$\sqrt{3}$. If this is wrong,then $z=\frac{6-3i}{\sqrt{15}}$ – Win_odd Dhamnekar Nov 11 '18 at 12:31
-
@DhamnekarWinod Exactly, we can obtain z form that. And you can check that |z| is indeed $\sqrt 3$. – user Nov 11 '18 at 12:39
1
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
lab bhattacharjee
- 274,582
-
-
https://artofproblemsolving.com/wiki/index.php?title=Without_loss_of_generality – lab bhattacharjee Nov 11 '18 at 13:03
-
I know $e^{i\pi}=-1$. So what is $e^{it}?$ and why did you multiply it by z? – Win_odd Dhamnekar Nov 11 '18 at 13:11
-
https://math.stackexchange.com/questions/2660361/how-do-i-interpret-eulers-formula – lab bhattacharjee Nov 11 '18 at 13:12