Find Principal Part Of $\frac{ze^{iz}}{(z^2+9)^2}$ Find the principal part  of $\frac{ze^{iz}}{(z^2+9)^2}$ at $z_0=3i$
Can I say that $$\frac{ze^{iz}}{(z^2+9)^2}=\frac{\frac{ze^{iz}}{(z+3i)^2}}{(z-3i)^2}$$
And look at $g(z)=\frac{ze^{iz}}{(z+3i)^2}$?
 A: Hint: Set $w=z-3i$ and then
$$\frac{ze^{iz}}{(z^2+9)^2}=\frac{(w+3i)e^{iw-3}}{w^4(1+\frac{6i}{w})^2}$$
with the expansions
$$e^u=1+u+\dfrac{u^2}{2!}+\dfrac{u^3}{3!}+\dfrac{u^2}{4!}+\cdots$$
and
$$\dfrac{1}{(1+u)^2}=1-2u+3u^2-\cdots$$
A: Note that both parts $\frac{1}{(z-3i)^2}$ as well as $g(z)$ contribute to  the  principal part of the function. 

We obtain
  \begin{align*}
\frac{ze^{iz}}{(z^2+9)^2}&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(z+3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i+z-3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i)^2\left(1+\frac{z-3i}{6i}\right)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{\left[3i+(z-3i)\right]e^{i(z-3i+3i)}}{-36}
\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\tag{1}\\
&=\left(-\frac{i}{12e^3(z-3i)^2}-\frac{1}{36e^3(z-3i)}\right)e^{i(z-3i)}\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\tag{2}\\
&=\left(-\frac{i}{12e^3(z-3i)^2}-\frac{1}{36e^3(z-3i)}\right)\\
&\qquad\qquad\cdot\left(1+i(z-3i)+\sum_{n=2}^\infty \frac{(i(z-3i))^n}{n!}\right)\\
&\qquad\qquad\cdot\left(1-\frac{1}{3i}(z-3i)+\sum_{n=2}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\right)\tag{3}\\
&=\color{blue}{-\frac{i}{12e^3(z-3i)^2}+\frac{1}{12e^3(z-3i)}}+\sum_{n=0}^\infty a_n(z-3i)^n\tag{4}
\end{align*}

Comment:


*

*In (1) we use the binomial series expansion and prepare the numerator for expansion at $z-3i$.

*In (2) we do some simplifications and separate the relevant terms with negative powers.

*In (3) we separate the constant and linear terms of the power series which contribute to the principal part.

*In (4) we multiply out and obtain the (blue colored) principal part.
