Polar form of a complex with square root

Question

Find the polar form of the roots of the polinomial

$p(z)=iz^2-2z+1+2i$

I found the roots $z_1=-i-\sqrt{i-3}$ and $z_2=-i+\sqrt{i-3}$ but I don't know how to deal with the complex numbers inside the square root in order to isolate both real and imaginary parts.

Answer 1

To compute the square root of a complex number: $(x+iy)^2=-3+i$, expand the l.h.s. and identify the real and imaginary parts: $$x^2-y^2=-3, \quad xy=\frac12.$$ You simplify the computation observing the square of the modulus of $x+iy$ is the modulus of $-3+i$: $$x^2+y^2=\sqrt{10}.$$ So we have a linear system in $x^2$ and $y^2$: $$\begin{cases}x^2+y^2=\sqrt{10}\\x^2-y^2=-3\end{cases}\iff\begin{cases}x^2=\dfrac{\sqrt{10}+3}2\\ y^2=\dfrac{\sqrt{10}-3}2\end{cases}$$ Observe $xy>0$, so $x$ and $y$ have the same sign, and ultimately $$x+iy=\pm\frac12\biggl(\sqrt{2\sqrt{10}+6}+i\sqrt{2\sqrt{10}-6}\biggr).$$

Answer 2

Given $y = a x^2+bx + c$, the roots are given by the quadratic formula $$ y_{\pm} = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}. $$ Here $$ a = i, \quad b = -2, \quad c = 1 + 2i, $$ therefore $$ z_{\pm} = -i \left(1 \pm \sqrt{3 - i} \right). $$ Compute the magnitude and argument.