Roots of a polynomial $f(x)$ in $\mathbb{C}[x]$

"Find all the roots of the polynomial $f(x)=x^2+(3i-2)x-2(1+i)$. Why does the answer not violate the $Conjugate \space Roots \space Theorem \space (CJRT)$"

I tried using the quadratic formula and got to $$x = \frac{-(3i-2) \pm \sqrt{3-4i}}{2}$$ but I'm having trouble with the $\sqrt{3-4i}$.

If I use de Moivre's Theorem I get $$\sqrt{5}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)$$ but I'm having trouble finding $\theta$

• Is that supposed to be $(3i-2)x$ in the polynomial? – KSmarts Dec 5 '14 at 22:56
• Yes it is, my bad. – Stephen Dec 5 '14 at 22:57

For your problem, to find the two values of $\sqrt{3 - 4 i}$, or of a general $\sqrt{x + i y}$, you have to solve the equation $(X+iY)^2 = x+iy$. This is equivalent to $X^2 - Y^2 = x$ and $2XY = y$. To which you can add $(X^2+Y^2)^2 = x^2+y^2$ by taking moduluses in the equation $(X+iY)^2 = x+iy$. So you have $X^2+Y^2 = \sqrt{x^2+y^2}$. Adding (resp. substracting) this to $X^2 - Y^2 = x$ gives you $X^2 = \frac{x + \sqrt{x^2+y^2}}{2}$ and $Y^2 = \frac{-x + \sqrt{x^2+y^2}}{2}$. Both previous right-hand sides (especially the second one) are always positive, so you'll find two (at most) values for $X$ and $Y$ (using the sign of $y$ through the $2XY = y$ equation) which finally give you roots (the plural handle one root with multiplicity $2$) of $x+iy$.
Application : $\sqrt{3 - 4 i} = \pm (2 - i)$.
Proof : $(\pm (2 - i))^2 = (2 - i)^2 = 4 - 1 - 2\times 2 i = 3 - 4 i$.
The point of the "conjugate roots theorem" is that if $P$ is a polynomial with real coefficients, then $P(\overline{z}) = \overline{P(z)}$; thus if $P(z_0) = 0$, then $P(\overline{z_0}) = \overline{P(z_0)} = 0$. Your given polynomial $f$ does not have real coefficients and do not satisfy $\overline{f(z)} = f(\overline{z})$.