Solved Problem on Cauchy's Integral Formula

f) \( \displaystyle \oint_{{|z|=1}}\frac{z^{2}}{z-2\mathrm{i}}\;dz \)

The path of integration is given by the circle of radius 1, centered at the origin (0, 0), traversed counterclockwise (Figure 1).
The general form of Cauchy's Integral Formula given by

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

Identifying the terms of the integral

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

Figure 1

The point \( z-2\mathrm{i}=0\Rightarrow z=2\mathrm{i} \) is outside the region determined by the closed contour C, using the Cauchy's Integral Theorem the integral is equal to zero.

\[ \begin{gather} \bbox[#FFCCCC,10px] {\oint_{{|z|=1}}\frac{z^{2}}{z-2\text{i}}\;dz=0} \end{gather} \]

Note 1: The path traversed \( |\;z\;|=1 \) is a circle. For a complex number \( z=x+\mathrm{i}y \), the absolute value is given by \( \sqrt{x^{2}+y^{2}\;}=1 \), squaring both sides of equation \( \left(\sqrt{x^{2}+y^{2}\;}\right)^{2}=1^{2} \), we obtain the equation of a circle \( (x-0)^{2}+(y-0)^{2}=1^{2}, \)

\[ (x-0)^{2}+(y-0)^{2}=1^{2} \]

with a radius equal to 1 and center at the origin (0, 0).

Note 2: f⁽⁰⁾ represents the calculation of the function at the point z₀ without derivative.

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