To solve the given problem, we need to first analyze and simplify the given complex integral and then determine [tex]\( f'(1+i) \)[/tex]. Here is the detailed, step-by-step solution:
1. Understanding the Integral:
We have the integral [tex]\( f(z) = \oint_C \frac{3z^2 + 7z + 1}{(z-1)^2} \, dz \)[/tex], where [tex]\( C \)[/tex] is a closed contour. The integral is of the form
[tex]\[
f(z) = \oint_C F(z) \, dz
\][/tex]
where [tex]\( F(z) = \frac{3z^2 + 7z + 1}{(z-1)^2} \)[/tex].
2. Check for Singularities:
The integrand [tex]\( F(z) \)[/tex] has a singularity at [tex]\( z = 1 \)[/tex], which is a pole of order 2 (since it's of the form [tex]\( \frac{\text{Polynomial}}{(z-1)^2} \)[/tex]).
3. Using the Residue Theorem:
According to the residue theorem, for a function with a simple pole, the integral around a closed contour enclosing the singularity is related to the residue of the pole. However, for a pole of order 2, we must use the following formula for the residue:
[tex]\[
\text{Res}(F(z), z=1) = \lim_{z \to 1} \frac{d}{dz} \left[ (z-1)^2 F(z) \right]
\][/tex]
4. Calculate the Residue:
Let's compute the residue at [tex]\( z = 1 \)[/tex]:
[tex]\[
(z-1)^2 F(z) = 3z^2 + 7z + 1
\][/tex]
Derive this with respect to [tex]\( z \)[/tex]:
[tex]\[
\frac{d}{dz} \left( 3z^2 + 7z + 1 \right) = 6z + 7
\][/tex]
Evaluate at [tex]\( z = 1 \)[/tex]:
[tex]\[
\left. (6z + 7) \right|_{z=1} = 6(1) + 7 = 13
\][/tex]
5. Conclusion for the Integral:
Thus, by the residue theorem, the integral simplifies to:
[tex]\[
f(z) = 2\pi i (\text{Residue at } z=1) = 2\pi i \cdot 13 = 26\pi i
\][/tex]
6. Determining [tex]\( f'(z) \)[/tex]:
Given [tex]\( f(z) \)[/tex] is a constant multiple of [tex]\( 2\pi i \)[/tex], it doesn’t change with [tex]\( z \)[/tex], and thus, [tex]\( f'(z) \)[/tex] must be 0 because the derivative of a constant is zero.
7. Evaluate [tex]\( f'(1+i) \)[/tex]:
Since the function [tex]\( f(z) = 26\pi i \)[/tex] is constant,
[tex]\[
f'(z) = 0
\][/tex]
Thus,
[tex]\[
f'(1+i) = 0
\][/tex]
Therefore, [tex]\( f'(1+i) = 0 \)[/tex].
