If $$f(a)=\oint_C\frac{3z^2+7z+1}{z-a}dz$$ where $C$ is the circle $|z|=2$, find
$(i)$ $f(1-i),$
$(ii)$ $f''(1-i),$
$(iii)$ $f(1+i).$
I am able to do first and third parts. Please help me with the second part.
What is the value of $f''(1-i)$.
Will it be zero or not exist since $1-i$ is a pole, or $12\pi i$?
On
Cauchy's Integral Formula states that if a complex function $f(z)$ is analytic in an open domain $S$, and if the closed disk $D=\{z:|z-z'|\le \rho\}$ is contained in $S$, then for any number $a$ inside $D$
$$f(a)=\frac{1}{2\pi i}\oint_{|z-z'|=\rho} \frac{f(z)}{z-a}\,dz \tag1$$
Note that $f(z)=3z^2+7z+1$ is an entire function and is therefore analytic in an on the closed disk $|z|\le 2$. Using $(1)$ with $f(z)=3z^2+7z+1$, $z'=0$, $\rho=2$, then for $|a|<2$ we have
$$3a^2+7a+1=\frac{1}{2\pi i}\oint_{|z|=2}\frac{3z^2+7z+1}{z-a}\,dz\tag 2$$
Moreover, Cauchy's Integral Theorem guarantees that for $|a|>2$
$$\oint_{|z|=2}\frac{3z^2+7z+1}{z-a}\,dz=0 \tag3$$
Putting $(2)$ and $(3)$ together yields
$$\oint_{|z|=2}\frac{3z^2+7z+1}{z-a}\,dz=\begin{cases}2\pi i (3a^2+7a+1)&,|a|<2\\\\0&,|a|>2\end{cases} $$
Therefore, we find that
$$f''(a)=\frac{d^2}{da^2}\left.\left(\oint_{|z|=2}\frac{3z^2+7z+1}{z-a}\,dz\right)\right|_{a=1-i} \begin{cases}12\pi i&,|a|<2\\\\0&,|a|>2\end{cases}$$
On
The suggestion given by Robert Israel is very good, of course. Another way of computing $f(a)$ is\begin{align*}f(a)&=\oint_C\frac{3z^2+7z+1}{z-a}\mathrm dz\\&=\oint_C\frac{3z^2+7z+1-(3a^2+7a+1)}{z-a}\mathrm dz+\oint_C\frac{3a^2+7a+1}{z-1}\mathrm dz\\&=\oint_C\frac{3(z-a)(z+a)+7(z-a)}{z-a}\mathrm dz+2\pi i(3a^2+7a+1)\operatorname{Ind}_C(a)\\&=\oint_C3(z+a)+7\,\mathrm dz+2\pi i(3a^2+7a+1)\text{ (if $|a|<2$)}\\&=2\pi i(3a^2+7a+1).\end{align*}Therefore, $f''\equiv12\pi i$ (in $D(0,2)$).
It looks like Cauchy's Integral formula... $$ f(a) = \frac{1}{2\pi i} \oint_C \frac{2\pi i(3z^2 + 7z+1)}{z-a}\,\mathrm dz = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-a}\;\mathrm dz$$ I'd say that $f(z) = 2\pi i(3z^2 + 7z+1)$... then
i) $f(1-i) = 26\pi + 16\pi i$
ii) $f''(1-i) = 12\pi i$
iii) $f(1+i) = -26\pi +16\pi i$