Solved Problem on Contour Integration
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c)   \( \displaystyle f(z)=\frac{z+2}{z} \),   along the circle   \( z=2\operatorname{e}^{\mathsf{i}\theta} \),   \( (0\leqslant \theta\leqslant 2\pi) \).

  • Parameterization of the γ curve, θ = t (Figure 1)
\[ \begin{gather} z(t)=2\operatorname{e}^{\mathsf{i}t} \end{gather} \]
Figure 1
  • Derivative of   \( z(t)=2\operatorname{e}^{\mathsf{i}t} \)
\[ \begin{gather} z'(t)=\frac dz dt=2\mathsf{i}\operatorname{e}^{\mathsf{i}t} \tag{I} \end{gather} \]
The contour integral is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\int f(z)\;dz=\int f(z(t))z'(t)\;dt} \tag{II} \end{gather} \]
  • Integration of   \( f(z)=\dfrac{z+2}{z} \)
\[ \begin{gather} f(z(t))=\frac{2\operatorname{e}^{\mathsf{i}t}+2}{2\operatorname{e}^{\mathsf{i}t}} \tag{III} \end{gather} \]
substituting expressions (I) and (III) into expression (II)
\[ \begin{split} \int f(z)\;dz &=\int_{0}^{2\pi}\;\left(\frac{\cancel{2}\operatorname{e}^{\mathsf{i}t}+\cancel{2}}{\cancel{2}\cancel{\operatorname{e}^{\mathsf{i}t}}}\right)\left(2\mathsf{i}\cancel{\operatorname{e}^{\mathsf{i}t}}\right)\;dt=\\[5pt] &=2\mathsf{i}\int_{0}^{2\pi}\;\operatorname{e}^{\mathsf{i}t}+1\;dt=\\[5pt] &=2\mathsf{i}\left[\int_{0}^{\pi}\;\operatorname{e}^{\mathsf{i}t}\;dt+\int_{0}^{2\pi}\;dt\right]=\\[5pt] &=2\mathsf{i}\left[\frac{1}{\mathsf{i}}\;\left.\operatorname{e}^{\mathsf{i}t}\;\right|_{0}^{2\pi}+\left.t\;\right|_{0}^{2\pi}\right]=\\[5pt] &=2\mathsf{i}\left[\frac{1}{\mathsf{i}}\times\frac{\mathsf{i}}{\mathsf{i}}\;\left(\operatorname{e}^{\mathsf{i}2\pi}-\operatorname{e}^{0}\right)+\left(2\pi-0\right)\right]=\\[5pt] &=2\mathsf{i}\left[\frac{\mathsf{i}}{\mathsf{i}^{2}}\;\left(\cancelto{1}{\cos 2\pi} +\mathsf{i}\cancelto{0}{\sin 2\pi} -1\right)+2\pi\right]=\\[5pt] &=2\mathsf{i}\left[\frac{\mathsf{i}}{-1}\;\left(1-1\right)2+\pi\right]=\\[5pt] &=2\mathsf{i}\left[-\mathsf{i}\;\left(0\right)+2\pi\right]=\\[5pt] &=4\pi\mathsf{i} \end{split} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\int_{0}^{2\pi}f(z)\;dz=4\pi} \end{gather} \]
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