Trigonometrical approach for negative discriminant of second order polynomial.

Question

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 1

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

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

Answer 2

The sign of the real component determines the sign of the cosine. The sign of the complex component determines the sign of the sine.

Answer 3

The complex number $a+bi\ne0$ can be uniquely represented as $$ a+bi=r(\cos\theta+i\sin\theta) $$ for a unique $r>0$ and a unique $\theta\in[0,2\pi)$. One can note that $$ r=\sqrt{a^2+b^2}=|a+bi| $$ and also that $a=r\cos\theta$, $b=r\sin\theta$. Then, if $a\ne0$, one can write $$ \tan\theta=\frac{b}{a} $$ but, unfortunately, this is not sufficient to determine $\theta$. But you don't need to use the arctangent: the identity above tells you that $$ \cos\theta=\frac{a}{r} \qquad \sin\theta=\frac{b}{r} $$ and this does uniquely determine $\theta\in[0,2\pi)$.

In your example of $a+bi=-8-6i$, you can compute $$ r=\sqrt{64+36}=10 $$ so you have $$ \cos\theta=-\frac{4}{5} \qquad \sin\theta=-\frac{3}{5} $$ This tells you that the angle is in the third quadrant, so it will be $$ \theta=\pi+\arcsin\frac{3}{5} $$

This doesn't really help in finding the square roots of $-8-6i$, which can be done directly if $$ -8-6i=(p+qi)^2=p^2-q^2+2pqi $$ then $$ \begin{cases} p^2-q^2=-8 \\[4px] 2pq=-6 \end{cases} $$ Therefore $q=-3/p$ and substituting in the first equation yields $$ p^2-\frac{9}{p^2}=-8 $$ that simplifies to $p^4+8p^2-9$. This is a biquadratic and we derive $$ p^2=\frac{-8+\sqrt{64+36}}{2}=\frac{-8+10}{2}=1 $$ Thus $p=1$ and $q=-3$ or $p=-1$ and $q=3$.

(The negative root of $t^2+8t-9=0$ should be obviously discarded.)