complex numbers - Trigonometrical approach for negative discriminant of second order polynomial.

This is in continuation to my earlier post, here

For trigonometrical approach to solve a complex second order polynomial (derived either directly, or resolved to be of that form) need consider discriminant. $D$ can be positive either directly, or if in the 3rd quad. can be converted as in second answer to my linked (earlier) post, to lie in the first.
But, for the cases where the discriminant is of mixed sign (say, $-8 +6i$ or $8-6i$), with angle in either 2nd or 4th quad. There it cannot be manipulated to represent angle in the first quadrant.

In 2nd quad., $\sin$ of negative angle is positive, while in the 3rd quad. both $\sin, \cos$ are negative.

But, further to that am hesitant to pursue.

Answer

Suppose we want to solve

$$z^2=-8+6i,$$

suppose we have found $z_1^2=-8+6i$, when we also have $(-z_1)^2=-8+6i$.

Write

$$-8+6i=r\exp(i\theta)$$

where $r$ is a positive number and $\theta$ is a real number.

$$r=\sqrt{6^2+8^2}=10.$$

That is we have

$$-8+6i=10\exp(i\theta)=10\cos(\theta)+10i\sin(\theta)$$

We have

$$\cos(\theta)=-\frac{8}{10}=-\frac{4}{5}$$
$$\sin(\theta)=\frac{6}{10}=\frac{3}{5}$$

Angle $\theta$ has negative cosine angle and positive sine angle, it is an angle in the second quadrant

$$2\cos^2\left( \frac{\theta}2\right)-1=-\frac45 \implies \cos^2\left( \frac{\theta}2\right)=\frac1{10}$$

$$2\sin\left( \frac{\theta}2\right) \cos\left(\frac{\theta}2 \right)=\frac3{5}$$

If $\cos\left(\frac{\theta}{2} \right)=\frac1{\sqrt{10}}$, then we have $\sin\left(\frac{\theta}2\right)=\frac{3}{{\sqrt{10}}}$.

If $\cos\left(\frac{\theta}{2} \right)=-\frac1{\sqrt{10}}$, then we have $\sin\left(\frac{\theta}2\right)=-\frac{3}{{\sqrt{10}}}$.

The square roots are

$$\pm10\left(\frac1{\sqrt{10}}+\frac{3i}{\sqrt{10}}\right)=\pm\sqrt{10}(1+3i)$$