Solved Problem on Cauchy's Integral Formula

d) \( \displaystyle \oint_{{C}}\frac{z^{2}}{z-1-\mathrm{i}}\;dz \)

The path is given by contour C (Figure 1). Point \( z-1-\mathrm{i}\Rightarrow z=1+\mathrm{i} \) is inside the region determined by C.
The Cauchy Integral Formula in the general form is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {f^{(n)}(z_{0})=\frac{{n}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz} \tag{III} \end{gather} \]

Identifying the terms of the integral

\[ \begin{gather} \frac{{ \bbox[#FFCC66,2px] {n} }!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{ \bbox[#FFFF66,2px] {f(z)} }{\left(z- \bbox[#FFD9CC,2px] {z_{0}} \right)^{ \bbox[#FFCC66,2px] {n} +1}}\;dz=\oint_{C}\frac{ \bbox[#FFFF66,2px] {z^{2}}} {\left[z- \bbox[#FFD9CC,2px] {(1+\mathrm{i})} \right]^{ \bbox[#FFCC66,2px] {0} +1}}\;dz \end{gather} \]

Figure 1

we have \( f(z)=z^{2} \), \( z=1+\mathrm{i} \) and n = 0, writing expression (III) for the integral

\[ \begin{gather} I_{1}=\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt] I_{1}=\oint_{{C}}\frac{z^{2}}{z-1-\mathrm{i}}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f^{(0)}(0)\\[5pt] I_{1}=2\pi\mathrm{i}(1+\mathrm{i})^{2}\;\\[5pt]I_{1}=2\pi\mathrm{i}(1+2\mathrm{i}+\mathrm{i}^{2})\\[5pt] I_{1}=2\pi\mathrm{i}(1+2\mathrm{i}-1)\\[5pt] I_{1}=2\pi \mathrm{i}(2\mathrm{i})\\[5pt] I_{1}=4\pi\mathrm{i}^{2}\\[5pt] I_{1}=4\pi \times(-1) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\oint_{{C}}\frac{z^{2}}{z-1-\mathrm{i}}\;dz=-4\pi} \end{gather} \]

Note 1: We do not need to know the equation that describes the contour C for the calculation, it is enough to know if the singularity points are inside or outside the region determined by the contour.
Note 2: f⁽⁰⁾ represents the calculation of the function at the point z₀ without derivative.

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