- State the Cauchy integral formula and use it to evaluate the complex line integral: \(\oint_\gamma z^2/(z-i)\, dz\), where \(\gamma\) is a circle centered at \(i\) and oriented counterclockwise.
- Use a parametrization of the path \(\gamma\) to express the line integral explicitly in terms of real integrals, with limits of integration. [You do not have to evaluate these real integrals.]
Solution
Cauchy’s integral formula states that if \(f\) is holomorphic on and inside a positively oriented simple closed contour \(\gamma\), and \(a\) lies inside \(\gamma\), then \[ \oint_\gamma \frac{f(z)}{z-a}\,dz=2\pi i\,f(a). \] Taking \(f(z)=z^2\) and \(a=i\) gives \[ \oint_\gamma \frac{z^2}{z-i}\,dz =2\pi i\,i^2 =-2\pi i. \]
Let the circle have radius \(R>0\). A counterclockwise parametrization is \[ z(t)=i+Re^{it},\qquad 0\le t\le2\pi, \] with \[ dz=iRe^{it}\,dt. \] Therefore \[ \oint_\gamma \frac{z^2}{z-i}\,dz =\int_0^{2\pi} i\bigl(i+Re^{it}\bigr)^2\,dt. \] Writing \(e^{it}=\cos t+i\sin t\), this becomes \[ \int_0^{2\pi} \Bigl[-2R\cos t\,(1+R\sin t) +i\bigl(R^2\cos^2t-(1+R\sin t)^2\bigr)\Bigr]dt, \] which is an explicit expression in real integrals with limits of integration.