1

**Quiz-6 / Re: Q6 TUT 0203**

« **on:**November 18, 2018, 02:17:49 AM »

This proof

Let $f(z)$ be a function analytic on the punctured disk of radius $R$ centered on $z_0$: $\{z | 0 < |z - z_0| < R\}$, and has a pole of order $l$ @ $z_0$.

Then, $\displaystyle f(z) = \frac{H(z)}{(z-z_0)^l}$, where $H$ is analytic and nonzero on

Therefore, we express $\displaystyle H(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + ... = \sum_{k=0}a_k(z-z_0)^k$

We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(z-z_0)^l - l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = \frac{H'(z)(z-z_0)^l}{(z-z_0)^{2l}} - \frac{l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = l\frac{H(z)}{(z-z_0)^{2l - l + 1}} - \frac{H'(z)}{(z-z_0)^{2l - l}}$

We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}$

It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(z-z_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)} - l\frac{1}{(z-z_0)^1}$

From now on, we can see that the residue could be $-l$, but considering $\frac{H'(z)}{1H(z)}$, we conclude that the $H$ being analytic and nonzero on the entire disk implies $H'$ being analytic as well, and finally $\frac{H'(z)}{H(z)}$, and so that fraction has

We conclude that the residue has to be $-l$.

**does not require**the direct evaluation of Laurent/Power series, but instead relies on algebraic and calculus manipulation.Let $f(z)$ be a function analytic on the punctured disk of radius $R$ centered on $z_0$: $\{z | 0 < |z - z_0| < R\}$, and has a pole of order $l$ @ $z_0$.

Then, $\displaystyle f(z) = \frac{H(z)}{(z-z_0)^l}$, where $H$ is analytic and nonzero on

*all*of the disk $\{z | 0 \leq |z - z_0| < R\}$.Therefore, we express $\displaystyle H(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + ... = \sum_{k=0}a_k(z-z_0)^k$

We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(z-z_0)^l - l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = \frac{H'(z)(z-z_0)^l}{(z-z_0)^{2l}} - \frac{l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = l\frac{H(z)}{(z-z_0)^{2l - l + 1}} - \frac{H'(z)}{(z-z_0)^{2l - l}}$

We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}$

It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(z-z_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)} - l\frac{1}{(z-z_0)^1}$

From now on, we can see that the residue could be $-l$, but considering $\frac{H'(z)}{1H(z)}$, we conclude that the $H$ being analytic and nonzero on the entire disk implies $H'$ being analytic as well, and finally $\frac{H'(z)}{H(z)}$, and so that fraction has

**no principal part**. This leaves the maximum negative degree of the principal part to be 1.We conclude that the residue has to be $-l$.