Solving a complex number equation

Question

I have this exercise:

Find the imaginary part of the solution whose real part is negative.

$z^2-2iz-\frac{1}{4}\left(5+i\sqrt{3}\right)=0$

I tried to factorize it and I came up with this:

$$\frac{8i\pm \sqrt{-16-4\times4\times(-5-i\sqrt3)}}{8}$$

Is it right? How can I complete the exercise?

Answer 1

By the quadratic formula, the roots of this equation are$$\frac{2i\pm\sqrt{1+\sqrt3i}}2=i\pm\frac12\sqrt{1+\sqrt3i}.$$But$$1+\sqrt3i=2\left(\frac12+\frac{\sqrt3}2i\right)=2\left(\cos\left(\frac\pi3\right)+\sin\left(\frac\pi3\right)i\right)$$and therefore the square roots of $1+\sqrt3i$ are$$\pm\sqrt2\left(\cos\left(\frac\pi6\right)+\sin\left(\frac\pi6\right)i\right)=\pm\sqrt2\left(\frac{\sqrt3}2+\frac i2\right).$$Therefore, the solution whose real part is negative is$$-\sqrt2\frac{\sqrt3}4+i-\sqrt2\frac i4=-\sqrt{\frac38}+i-\frac i{\sqrt8}.$$

Answer 2

No, no, no. You must write $z=a+bi$; $a,b\in\mathbb{R}$ and $a<0$. Put this into equation and do some calculation.