Question:
Let $z = 2 + \sqrt{2} - (3 + 3 \sqrt{2})i$, and let $c = 2 - 3i$.  Let $w$ be the result when $z$ is rotated around $c$ by $\frac{\pi}{4}$ counter-clockwise.

[asy]
unitsize(0.6 cm);

pair C, W, Z;

Z = (2 + sqrt(2), -3 - 3*sqrt(2));
C = (2,-3);
W = rotate(45,C)*(Z);

draw(Z--C--W);

dot("$c$", C, N);
dot("$w$", W, SE);
dot("$z$", Z, S);
label("$\frac{\pi}{4}$", C + (0.6,-1));
[/asy]

Find $w.$

Answer:
Rotating by $\frac{\pi}{4}$ counter-clockwise corresponds to the complex number
\[e^{\pi i/4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}.\]Hence,
\[w - c = \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (z - c),\]so
\begin{align*}
w &= \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (z - c) + c \\
&= \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (\sqrt{2} - 3i \sqrt{2}) + 2 - 3i \\
&= (4 - 2i) + 2 - 3i \\
&= \boxed{6 - 5i}.
\end{align*}