Quadratic equation with complex number as a parameter

Question

I used the formula for a quadratic equation, but the square root was problematic to solve this problem. It supposed to be done without trigonometry. $$z^2-(2+i)z-1+7i=0$$

Answer 1

HINT

We can simply apply the quadratic formula which holds also for complex numbers

$$az^2+bz+c=0\implies z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

then

$$z=\frac{(2+i)\pm\sqrt{(2+i)^2-4(-1+7i)}}{2}$$

For the calculation of the square roots we can use trigonometric form

$$w=r(\cos \theta + i \sin \theta) \implies \sqrt w=\sqrt r\left(\cos \left(\frac{\theta}2+k\pi\right) + i \sin \left(\frac{\theta}2+k\pi\right)\right) \quad k=0,1 $$

Answer 2

Use Quadratic Formula to get:

$$z=\frac{(2+i)\pm\sqrt{7-24i}}{2}$$

Then use the following result: $$\text{Let}\space (a+bi)^2=c+di$$ $$\text{Then}\space a=\pm\sqrt{\frac{c\pm\sqrt{c^2+d^2}}{2}}$$ $$\text{And}\space b=\pm\sqrt{\frac{-c\pm\sqrt{c^2+d^2}}{2}}$$

Checking the signs as you go.

This result can be proven without the use of trigonometry (and indeed, I did not know about polar forms when I found it)