Solved Problem on Contour Integration

a) \( f(z)=z \), along the path from the origin to point 1+i.

Parameterization of the γ curve (Figure 1)

The parameterization of a line is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {z(t)=z_{1}+t(z_{2}-z_{1})} \end{gather} \]

where z₁ and z₂ are the start and end points of the curve, \( z_{1}=0+0i \) and \( z_{2}=1+i \)

\[ \begin{gather} z(t)=(0+0i)+t[(1+i)-(0+0i)]\\[5pt] z(t)=t[1+i]\ \ \phantom{{}}\\[5pt] \qquad \qquad \qquad \qquad z(t)=t+it\qquad ,\qquad 0\leqslant t\leqslant 1 \tag{I} \end{gather} \]

Figure 1

Derivative of z(t) = t+it

\[ \begin{gather} z'(t)=\frac{dz}{dt}=1+i \tag{II} \end{gather} \]

The contour integral is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\int f(z)\;dz=\int f(z(t))z'(t)\;dt} \tag{III} \end{gather} \]

Integration of f(z) = z

\[ \begin{gather} f(z)=z=x+iy \tag{IV} \end{gather} \]

using expression (I) we have f(z(t))

\[ \begin{gather} z(t)=\underbrace{\ t\ }_{x(t)}+\underbrace{\ t\ }_{y(t)}i \tag{V} \end{gather} \]

substituting the values of x(t) and y(t) of expression (V) into expression (IV)

\[ \begin{gather} f(z(t))=t+it \tag{VI} \end{gather} \]

substituting expressions (II) and (VI) into expression (III)

\[ \begin{split} \int f(z)\;dz &=\int_{0}^{1}(t+it)(1+i)\;dt=\\[5pt] &=\int_{0}^{1}t(1+i)(1+i)\;dt=\\[5pt] &=\int_{0}^{1}t(1+i)^{2}\;dt=\\[5pt] &=(1+i)^{2}\int_{0}^{1}t\;dt=\\[5pt] &=(1^{2}+2i+i^{2})\;\left(\left.\frac{t^{2}}{2}\;\right|_{0}^{1}\right)=\\[5pt] &=(1+2i-1)\;\left(\frac{1}{2}\right)=\\[5pt] &=\cancel{2} i\times\frac{1}{\cancel{2}} \end{split} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\int_{0}^{1+i}f(z)\;dz=i} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .