\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Solve the equation $$ z^2 = -5 +2i.$$  Express the solution in the
form $a+ib,\, \, $ $a,b \epsilon {\bf R}$

\vspace{.25in}

{\bf Answer}

$ z^2 = -5 +2i. \, \, a,b \epsilon {\bf R}$

$\ds a^2 - b^2 = -5 \Rightarrow 2ab = 2 \Rightarrow b =
\frac{1}{a}$

$\ds a^2 - \frac{1}{a^2} = -5 \Rightarrow a^4 + 5a^2 - 1 = 0
\Rightarrow a^2 = \frac{-5\pm \sqrt{29}}{2}$ and $a \epsilon {\bf
R}$

${}$

So $\ds a^2 = \frac{-5+\sqrt {29}}{2} \Rightarrow a = \pm
\sqrt{\frac{\sqrt{29}-5}{2}} \Rightarrow b = \pm
\sqrt{\frac{\sqrt{29}+5}{2}}$

${}$

$$z = \left(\sqrt{\frac{\sqrt{29}-5}{2}} + i
\sqrt{\frac{\sqrt{29}+5}{2}}\right)$$

\end{document}
