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\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 239/17:}
\begin{description}
\item[a)]
The range is all $\R^2$ except $(0,0)$.  Given $(s,t)\in\R^2$, just
let $x = \frac12\log(s^2+t^2)$ and choose $y$ so that $\cos y = e^{-x}s$
and $\sin y = e^{-x}t$.  This can be done since 
$(e^{-x} s)^2 + (e^{-x} t)^2) = 1$. 
\item[b)] 
The Jacobian of $f$ is 
$$
\det\left(\begin{array}{cc}
e^x \cos y & -e^x\sin y \\
e^x \sin y & e^x \cos y\end{array}\right)
= 
e^{2x}(\cos^2 y + \sin^2 y) = e^{2x} \neq 0
$$
for any $(x,y)\in\R^2$.  So $f$ is locally invertible by the inverse
function theorem.  However, $f$ is not globally invertible since it's
not injective: $f(x,y+2n\pi) = f(x,y)$ for every $n\in\N$.
\item[c)]
$$
g(s,t) = (\frac12\log(s^2 + t^2),\tan^{-1}(t/s)),
$$
(choosing $\tan^{-1}$ to have range $(-\pi/2,\pi/2)$).  Then
$$
g'(s,t) = \left(\begin{array}{cc}
\frac{s}{s^2 + t^2} & \frac{t}{s^2+t^2} \\
\frac{-t}{s^2 + t^2} & \frac{s}{s^2+t^2}
\end{array}\right).
$$
So 
$$
g'(f(x,y)) = \left(\begin{array}{cc}
e^{-x}\cos y & e^{-x}\sin y \\
-e^{-x}\sin y & e^{-x}\cos y
\end{array}\right) = f'(x)^{-1}.
$$

\item[d)] 
For given $y$, the set
$$
\{f(x,y):x\in\R\} = \{e^x(\cos y,\sin y):x\in\R\}
$$
consists of all positive multiples of the fixed vector $(\cos y,\sin y)$.  
That is, the image of a horizontal line is a ray beginning at $(0,0)$ (but
not including this point).

For given $x$, the set
$$
\{f(x,y):y\in\R\} = \{e^x(\cos y,\sin y):y\in\R\}
$$
consists of all points at distance $e^x$ from $(0,0)$.  That is, the 
image of a vertical line is a circle centered at the origin.
\end{description}
\end{document}