\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION


\begin{description}

\item[(a)]
Let $a$ denote a real number, where $-1<a<1$. Derive the Laurent
expansion $${a\over z-a}=\sum_{n=1}^\infty{ a^n\over z^n}$$ that
is valid for $|a|<|z|<\infty.$

\item[(b)] By writing $z=e^{i\theta}$ and equating real and imaginary
parts, use the result in part (a) to derive the formulae
$$\sum_{n=1}^\infty a^n\cos n\theta = {a\cos\theta-a^2\over
1-2a\cos \theta +a^2},$$ and $$\sum_{n=1}^\infty a^n\sin
n\theta={a\sin\theta\over 1-2a\cos \theta +a^2}.$$

\end{description}



ANSWER


\begin{description}

\item[(a)]
${a\over z-a}={a\over z}(1-{a\over z})^{-1}$. As $|a|<|z|$, we can
write this as ${a\over z}(1+{a\over z}+{a\over
z}^2+\cdots)=\sum_{n=1}^{\infty}{a^n\over z^n}$.

\item[(b)]
Just put $e^{i\theta}=\cos \theta+ii\sin\theta$, use De Moivre's
Theorem and
 equate real and imaginary parts.

\end{description}



\end{document}
