Quadratic equation with "i" in discriminant

Question

I'm solving some complex problems (pun intended), but I'm unable to solve any of this one type. One just has to solve a quadratic equation, but an imaginary number is in the discriminant. So to give an example:

$z \bar z - 3z + 3 + i = 0$

So I calculated the discriminant $-3 - 4i$ and now I proceed to calculate roots of such number to plug them into the formula for "normal" roots. But I cannot solve $\arctan(\frac43)$. The frustrating part is that the solutions are quite easy to come by using factorization, but I want to know what is the mistake I'm making.

Any hints?

Answer 1

You properly computed the discriminant to be $-3-4 i$. Now, what you need it to compute the square root of this number, that is to say to find a complex number $x=a+i b$ such that $x^2=-3-4i$.

So, develop $$x^2=(a+i b)^2=a^2+2a i b -b^2$$ and identify the real and imaginary parts. This gives you $$a^2-b^2=-3$$ $$2 a b = -4$$ Extract $b=- \frac {2}{a}$ from the last equation and replace in the first. This gives you $$a^2-\frac {4}{a^2}=-3$$ you can easily transform in a quadratic equation in $a^2$. Fortunately, you have the obvious solution $a=1$ and then $b=-2$. So $$\sqrt {-3-4i}=1-2i$$

I am sure that you can take from here and finish your problem.

Answer 2

$\arctan(4/3)$ is a perfectly good solution to $\arctan(4/3)$. You can leave your answer in terms of it, but really what you ultimately want to do here is to simplify expressions like $\sin \arctan(4/3)$. That you can do easily: draw the relevant a triangle. (and make sure you pay attention to the fact that there are two different angular positions that have tangent $4/3$)