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\noindent{\Large{\textsc{Complex Variables\hfill Assignment 3.1  \\ Spring 2020 \hfill Due February 12}}}

\large

\vspace{1cm}


\begin{exercise}
Show that for $z \in \C$ the principal value of $\sqrt{z}$ is given by
\[
 \sqrt{\frac{|z| + \operatorname{Re}z}{2}} + i \operatorname{sgn}(\operatorname{Im} z) \sqrt{\frac{|z| - \operatorname{Re}z}{2}},
\]
where $\operatorname{sgn}(\cdot)$ is the sign function.  
\end{exercise}

\begin{exercise}
Give necessary and sufficient conditions on $z, z_1, z_2 \in \C$ so that the following equations hold.

\begin{itemize}

\item[\bf a.]  $\sqrt{z_1 z_2} = \sqrt{z_1} \sqrt{z_2}$.

\item[\bf b.]  $\sqrt{z^2} = z$.

\end{itemize}

\end{exercise}

\begin{exercise}
Let $a \in \R^+$.  For $z \in \C\setminus \! \{ \pm a\}$, let 
\[
f(z) = \sqrt{z+a}\sqrt{z-a}.
\]
Show that $f(z)$ defines a branch of $\sqrt{z^2 - a^2}$ that is continuous on $\Omega = \C \setminus [-a,a]$.  Determine $f(x+i0)$ and $f(x-i0)$ for $x \in (-a,a)$.  [{\em Suggestion:}  In light of what we know about the function $\sqrt{z}$, it suffices to show that $f(x+i0) = f(x-i0)$ for $x \in (-\infty, -a)$.]

\end{exercise}

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