$\frac{1}{z^2}$ is holomorphic

Question

I have to show that $z\mapsto\frac1{z^2}$ is holomorpic on $\mathbb C\setminus\{0\}$ and compute its $n$-th derivative

I know that $\frac{1}{z^2}=\sum\limits_{n\ge0}(-1)^n(n+1)(z-1)^n$, so it has a power series representation. Is that sufficient to conclude that the function is analytic, which is stronger than holomorphic. I thought analytic means inifinitely differentiable and holomorphic only continuously differentiable, how can one then compute $n$-th derivative of a holomorphic function, if $n$ is greater than $1$? However I computed it;

$\left(\frac{1}{z^2}\right)^{(n)}=\sum\limits_{k\ge n}(-1)^k(n+1)n!(z-1)^{k-n}$ is that correct ?

Answer 1

Given$$f(z) = {z^{ - 2}}$$ then, for$$z = x + iy$$ we may write$$f(x + iy) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} + i\frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}}$$ For$$\begin{gathered} u(x,y) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} \hfill \\ v(x,y) = \frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}} \hfill \\ \end{gathered}$$ Cauchy-Riemann Equations are satisfied.

For n-th order derivative, we get for $n \in \mathbb{N}$: $$\frac{{{d^n}f}}{{d{z^n}}}(z) = \left\{ {\begin{array}{*{20}{c}} {{{( - 1)}^{2n - 1}}(2n)! \cdot z \cdot f{{(z)}^{2n - 1}}} \\ {{{( - 1)}^{2n}}(2n + 1)!f{{(z)}^{2n - 1}}} \end{array}} \right.$$

One gets this by carefully calculating: $$\begin{gathered} \frac{{df}}{{d{z^1}}}(z) = {( - 1)^1} \cdot 2! \cdot z \cdot f{(z)^2} \hfill \\ \frac{{{d^3}f}}{{d{z^3}}}(z) = {( - 1)^3} \cdot 4! \cdot z \cdot f{(z)^3} \hfill \\ \frac{{{d^5}f}}{{d{z^5}}}(z) = {( - 1)^5} \cdot 6! \cdot z \cdot f{(z)^4} \hfill \\ \vdots \hfill \\ \hfill \\ \frac{{{d^2}f}}{{d{z^2}}}(z) = 3! \cdot f{(z)^2} \hfill \\ \frac{{{d^4}f}}{{d{z^4}}}(z) = 5! \cdot f{(z)^3} \hfill \\ \frac{{{d^6}f}}{{d{z^6}}}(z) = 7! \cdot f{(z)^4} \hfill \\ \vdots \hfill \\ \end{gathered}$$

Answer 2

If we know that $f(z)$ and $g(z)$ have complex derivatives at $z=z_0$, then $$ \begin{align} &\lim_{z\to z_0}\frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}\\ &=\lim_{z\to z_0}\frac{f(z)g(z)-f(z)g(z_0)+f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}\\ &=\lim_{z\to z_0}\frac{f(z)g(z)-f(z)g(z_0)}{z-z_0} +\lim_{z\to z_0}\frac{f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}\\ &=\lim_{z\to z_0}f(z)\frac{g(z)-g(z_0)}{z-z_0} +g(z_0)\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\\[6pt] &=f(z_0)g'(z_0)+f'(z_0)g(z_0)\tag{1} \end{align} $$

Using $(1)$, $\frac{\mathrm{d}}{\mathrm{d}z}1=\lim\limits_{z\to z_0}\frac{1-1}{z-z_0}=0$, $\frac{\mathrm{d}}{\mathrm{d}z}z=\lim\limits_{z\to z_0}\frac{z-z_0}{z-z_0}=1$, and induction, we get for $n\ge0$ $$ \frac{\mathrm{d}}{\mathrm{d}z}z^n=nz^{n-1}\tag{2} $$ If we know that $f(z)$ has a complex derivative at $z=z_0$ and that $f(z_0)\ne0$, then $$ \begin{align} \lim_{z\to z_0}\frac{\frac1{f(z)}-\frac1{f(z_0)}}{z-z_0} &=\lim_{z\to z_0}\left[-\frac1{f(z)f(z_0)}\frac{f(z)-f(z_0)}{z-z_0}\right]\\ &=-\frac1{f(z_0)^2}f'(z_0)\tag{3} \end{align} $$ Thus, $1/f(z)$ also has a complex derivative at $z=z_0$.

Using $(2)$ and $(3)$, we see that we can extend $(2)$ to $n\in\mathbb{Z}$: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}z}z^{-n} &=-\frac1{z^{2n}}nz^{n-1}\\ &=-nz^{-n-1}\tag{4} \end{align} $$ Equation $(4)$ for $n=2$, says that $z^{-2}$ has a complex derivative for $z\ne0$; therefore, it is holomorphic on $\mathbb{C}\setminus\{0\}$.

Furthermore, by induction on $(4)$, we get $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}z}z^{-2}&=-2z^{-3}\\ \frac{\mathrm{d}^2}{\mathrm{d}z^2}z^{-2}&=6z^{-4}\\ \frac{\mathrm{d}^3}{\mathrm{d}z^2}z^{-2}&=-24z^{-5}\\ &\vdots\\ \frac{\mathrm{d}^n}{\mathrm{d}z^n}z^{-2}&=(-1)^n(n+1)!\,z^{-n-2}\tag{5} \end{align} $$