Question

The vlaue of ${\oint }_C\frac{1}{z^2}dz$ where the contour is the unit circle traversed clockwise is

• A. $-2\pi i$
• B. $0$
• C. $-2\pi i$
• D. $4\pi i$

Answer 1

The correct option is B $0$

Method I:

Let $f\left(z\right)=\frac{1}{z^2}$ so poles are $z=0,0$ (Double pole)
and $c|z|=1|$ so pole lies inside ${}^{\prime }{c}^{\prime }$
$={\left[\frac{d}{dz}\right(1\left)\right]}_{z=0}=0$
By C-R-T,
${\oint }_{c}f\left(z\right)dz=2\pi \left(R\right)=2\pi i\left(0\right)=0$

Method II:
$\because$ Cauchy integral formula for $I^{st}$ derivative is
$f^{\prime }\left(Z_o\right)=\frac{1}{2\pi i}{\oint }_c\frac{f\left(z\right)dz}{(z-z_0{)}^2}$
So ${\oint }_c\frac{1}{Z^2}{d}_z={\oint }_C\frac{1}{(Z-0{)}^2}dz=2\pi i{f}^{\prime }\left(0\right)=0$