Question

The value of ${\int }_c\frac{z^2}{z^4-1}dz$, using Cauchy's integral formula, around the circle $|z+1|=1$ where $z=x+iy$ is

• A. $2\pi i$
• B. $-\pi i/2$
• C. $-3\pi i/2$
• D. ${\pi }^{2}i$

Answer 1

The correct option is B $-\pi i/2$

Method I:
Using Cauchy's residue theorem
Let $f\left(z\right)=\frac{z^2}{z^4-1}$
so pole of $f\left(z\right)$ are ; $z^4-1=0$
$(z^4-1)(z^2+1)=0$
$\Rightarrow z=\pm 1$ and $\pm i$
i.e., there exist four poles $1,-1,i,-i$ and all are simple poles.
Now given contour is $c:|z+1|=1$ (circle)
with centre at $z_0=-1=(-1,0)$ and raddius $=1$
So only $z=-1$ lies inside ${}^{\prime }{c}^{\prime }$
$R=\underset{(z=-1)}{Resf\left(z\right)}=\underset{z\to -1}{lim}(z+1)f\left(z\right)$

$={lim}_{z\to -1}\left[\right(z+1\left)\frac{z^2}{(z^4-1)}\right]$ $\left(\frac{0}{0}\text{from}\right)$
$=={lim}_{z\to -1}\left(\frac{3z^2+2z}{4z^3}\right)=-\frac{1}{4}$
Hence by Cauch's residue theorem.
${\oint }_{c}f\left(z\right)dz=2\pi i\left[R\right]=-\frac{\pi i}{2}$

Method II:
$\int \frac{Z^2}{Z^4-1}dz=\int \frac{Z^2}{(Z^2-1)(Z^2+1)}dz$
$\frac{1}{2}{\int }_c(\frac{1}{(Z^2-1)}+\frac{1}{(Z^2+1)})dz$
$I=\frac{1}{2}\int \frac{1}{Z^2-1}+\frac{1}{2}\oint \frac{1}{Z^2+1}dz$
$Cis|Z+1|=1$
$=\frac{1}{2}\oint =\frac{1}{(Z-1)(Z+1)}dz+=\frac{1}{2}\int =\frac{1}{(Z+i)(z-i)}dz$
$=\frac{1}{2}{\int }_c\frac{\left(\frac{1}{Z_1}\right)}{(Z+1)}dz+0$
[$\because$ Only $z=-1$ lies inside $C$.]
So 2nd integral becomes analytic and its's integral will be zero using cauchy integral theorem.
Now by Cauchy Integral formula,
$I=\frac{1}{2}2\pi i{\left[\frac{1}{Z-1}\right]}_{z=-1}$
$I=-\frac{\pi i}{2}$
Note: We can also use cauchy Residue theorem to evaluate above question.