Question

In the following integral, th contour $C$ encloses the points $2\pi j$ and $-2\pi j$.
$-\frac{1}{2\pi }{\oint }_C\frac{sinz}{(z-2\pi j{)}^3}dz$
The value of the integral is

• A. 133.87

Answer 1

The correct option is A 133.87

Method I:

Let $f\left(z\right)=\frac{sinz}{(z-2\pi j{)}^2}$
So poles are $z=2\pi j$ (poles of order $3$)
$R=\underset{(z=2\pi j)}{Resf\left(z\right)}$Diagram in equation

$(\because sin\theta =\frac{e^{i\theta }-e^{-i\theta }}{2i})$
$=\frac{e^{i\pi }-e^{-i\pi }}{-4j}$
$\because$ Pole lies inside${}^{\prime }{c}^{\prime }$ so by C - R - T
$-\frac{1}{2\pi }\oint \frac{sinz}{(z-2\pi j{)}^3}dz=-\frac{1}{2\pi }{\oint }_{c}f\left(z\right)dz$
$=-\frac{1}{2\pi }\left[2\pi i\right(R\left)\right]$
$=-\frac{1}{2\pi }\left[2\pi i\left(\frac{e^{-2\pi }-e^{2\pi }}{-4j}\right)\right]=\frac{e^{-2\pi }-e^{2\pi }}{4}$
$=-133.87$
Method II:
$I=-\frac{1}{2\pi }{\oint }_C\frac{sinz}{(z-j2\pi {)}^3}dz$
Using cauchy's integral formula;
$I=-\frac{1}{2\pi }\times j2\pi \frac{f^{\prime \prime }\left(j2\pi \right)}{2!}$
Let $f\left(z\right)=sinz$
$f^{\prime }\left(z\right)=cosz$
$f^{\prime \prime }\left(z\right)=-sinz$
$\therefore$ at $z=j2\pi ;f^{\prime }(z=j2\pi )=-sin\left(j2\pi \right)$
$\Rightarrow I=-\frac{1}{2\pi }\times j2\pi \times \frac{-sin\left(j2\pi \right)}{2!}$
$=-133.87$