<s>[INST] How to determine if $f(z)=\frac{i}{z^8}$ is analytic? How can I prove that a function like $$f(z)=\dfrac{i}{z^8}$$ is analytic or not?
I have to use Cauchy–Riemann equations but I can't find the $u$ and $v$ functions.
 [/INST] First, remember how complex division works:
$$
\frac{1}{w}=\frac{\overline{w}}{\lvert w\rvert ^2}.
$$
Also, recall that complex conjugates break up over multiplication: $\overline{uw}=\bar{u}\bar{w}$.
So, in all, if $z=x+iy$, then
$$
\frac{i}{z^8}=i\left(\frac{\overline{z^8}}{\lvert z^8\rvert ^2}\right)=i\left(\frac{(\bar{z})^8}{\lvert z\rvert^{16}}\right)=i\frac{(x-iy)^8}{(x^2+y^2)^8}.
$$
Now, by the Binomial Theorem,
$$
(x-iy)^8=\sum_{n=0}^{8}\binom{8}{n}x^{8-n}(-iy)^{n},
$$
from which we get
$$
\begin{align*}
\Re[(x-iy)^8]&=x^8-28x^6y^2+70x^4y^4-28x^2y^6+y^8,\\
\Im[(x-iy)^8]&=-8x^7y+56x^5y^3-56x^3y^5+8xy^7.
\end{align*}
$$
So, in all, you want to take the following for $u$ and $v$:
$$
\begin{align*}
u(x,y)&:=\Re\left[i\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=-\Im\left[\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\frac{8x^7y-56x^5y^3+56x^3y^5-8xy^7}{(x^2+y^2)^8}
\end{align*}
$$
and
$$
\begin{align*}
v(x,y)&:=\Im\left[i\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\Re\left[\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\frac{x^8-28x^6y^2+70x^4y^4-28x^2y^6+y^8}{(x^2+y^2)^8}.
\end{align*}
$$
</s>