Finding $z$ satisfying $2z=(3+4i)\bar z$, $z\bar z=2$

Question

Find the complex number $z$ satisfying both equations: $$5z=(3+4i)\bar z\\z\bar z=5$$

sorry, question has been revised..

Answer 1

Let $z=x+yi,$ where $x$ and $y$ are reals. Thus, $$5(x+yi)=(3+4i)(x-yi),$$ which gives $x=2y$.

Also, we have $x^2+y^2=5.$

Can you end it now?

Answer 2

You can use both rectangular or polar representation of the complex number to solve this. $$5z=(3+4i)\bar z$$ $$z\bar z=5$$ We know that : $$z\bar z=|z|^2=5$$ so, $$|z|=\sqrt 5$$

Put $z=|z|e^{i\theta}=\sqrt5 e^{i\theta}$ in the first equation,

$$5\sqrt5e^{i\theta}=(3+4i)\sqrt5e^{-i\theta}$$ $$e^{2i\theta}=\frac{(3+4i)}{5}$$ Now, $$\cos{2\theta}+i\sin{2\theta}=\frac{3}{5}+\frac{4i}{5}$$ Solve, $$\cos {2\theta}=\frac{3}{5}$$ and $$\sin {2\theta}=\frac{4}{5}$$ simultaneously to get the value of $\theta$.

Final answer is: $z=\sqrt5 e^{i\theta}$ where $\theta=\frac{1}{2}\tan^{-1} \frac{4}{3}$ Hope this helps...