complex induction zero points holomorphic function

Let $h:G\rightarrow \mathbb{C}$ be a holomorph function and $z_{0}$ a point in $G$ with $h(z_{0}) \ne 0$ . Let m be a natural number. Show that the function : $f:G\rightarrow \mathbb{C}$ defined by $f(z)=(z-z_{0})^{m}h(z)$ at $z_{0}$ ist a zero point of order m and that it holds that : $f^{(m)}(z_{0}) = m!h(z_{0})$

This is my proof:

Induction hypothesis : $f^{(m)}(z_{0}) = m!h(z_{0})$

Induction beginning : $m=0 \rightarrow f^{(0)}(z_{0}) = h(z_{0})$

Induction step : $m \rightarrow m+1$: $f^{(m+1)}(z_{0})= ((z-z_{0})^{m+1}h(z_{0}))^{(m+1)} = h(z_{0})^{(m+1)}(z_{0}-z_{0})^{m+1} + (m+1)!h(z_{0}) = (m+1)!h(z_{0})$

Is this proof correct. Tell me please.

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What do you mean by "doesn[']t look right"? – Henning Makholm Nov 6 '11 at 14:44
the solution has another way of doing it…. So i think this is not correct but I don't see why ! – VVV Nov 6 '11 at 14:51
What makes you think there cannot be two ways of doing it that are both correct? Solution keys are just for showing one way of doing it. If your proof works, then it's good. – Henning Makholm Nov 6 '11 at 14:53
In the past I often thougth I had proved something by induction where in fact I just restated the hypothesis. Is this the case here too? – VVV Nov 6 '11 at 14:55
I don't understand your last comment. I would write $f(z)=(z-z_o)^{m+1}.h(z)$, take the derivative of both sides: $f'(z)=I+II$, and apply the induction hypothesis at level $m$ to $I$ and $II$. The point to keep in mind is that $f^{(m+1)}(z)=(f')^{(m)}(z)$. – Georges Elencwajg Nov 6 '11 at 19:36

Another way to prove it is to use Cauchy integral formula: since $h(z)$ is holomorphic in $G$, $f(z)=(z-z_{0})^{m}h(z)$ is also holomorphic in $G$. For $r>0$ sufficiently small such that $\{z:|z-z_0|\leq r\}\subset G$, by Cauchy integral formula we have $$f^{(m-1)}(z_0)=\frac{(m-1)!}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{m}}dz=\frac{(m-1)!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)(z-z_{0})^{m}}{(z-z_0)^{m}}d z$$$$=\frac{m!}{2\pi i}\int_{|z-z_0|=r}h(z)dz=0.$$ Here the last equality follows from the fact that $h(z)$ is holomorphic. By Cauchy integral formula once again, we have $$f^{(m)}(z_0)=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{m+1}}dz=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)(z-z_{0})^{m}}{(z-z_0)^{m+1}}d z$$$$=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)}{z-z_0}dz=m!h(z_0)\neq0.$$ This shows that $z_0$ is a zero of order $m$ to $f(z)$, and $f^{(m)}(z_0)=m!h(z_0)$, as required.

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It would be best to include your definition of the order of a zero, because this essentially is the definition according to some. Alternatively, $f$ has a zero of order $m$ at $z_0$ if $f^{(k)}(z_0)=0$ when $k\in\{0,1,\ldots,m-1\}$, and $f^{(m)}(z_0)\neq 0$. You can find $f^{(k)}(z)$ using the product rule. First you could note that when $0\leq k\leq m$, the $k^\text{th}$ derivative of $(z-z_0)^m$ is $\frac{m!}{(m-k)!}(z-z_0)^{m-k}$. Therefore
$$f^{(k)}(z)=\sum_{n=0}^k{k \choose n}\frac{m!}{(m-k+n)!}(z-z_0)^{m-k+n}h^{(n)}(z).$$
When $k<m$, the exponent of $(z-z_0)$ is positive in each summand, so $f^{k}(z_0)=0$. When $k=m$, the only summand without a positive exponent on $(z-z_0)$ is where $n=0$, so we get $f^{(m)}(z_0)={m \choose 0}\frac{m!}{0!}h(z_0)=m!h(z_0).$ Since $h(z_0)\neq 0$, this also shows that the zero is of order $m$.
Although no explicit induction appears in this answer, both the product rule for higher derivatives and the formula for the higher derivatives of $(z-z_0)^m$ could be proven by induction.
$_{\text{(For simplicity I am using the notational convention }(\text{anything})^0=1\text{.)}}$