Question

Find the residues of \( f(z)=\frac{sin \pi z^2+coz \pi z^2}{(z-1)(z-2)^2} \) at its poles.

Answer 1

Solution:

Let,

\( \begin{aligned}I=\oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z \end{aligned} \)

Poles of the integrand are given by putting the denominator equal to zero.

\( (z-1)(z-2)^2=0 \Rightarrow z=1,2 \)

The integrand has two poles at z = 1, 2

both the poles z = 1, and z = 2.

\( \begin{aligned}I &= \oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z\\ & =\oint \frac{f(z)dz}{(z-1)(z-2)^2} \end{aligned} \)

 

Using partial fraction method, 

\( \begin{aligned} \frac{1}{(z-1)(z-2)^2} =\frac{A}{(z-1)}+\frac{B}{(z-2)} &+\frac{C}{(z-2)^2} \end{aligned} \)

 

By solving the partial fraction, we get

\( A=-1,B=\frac{1}{2},C=\frac{1}{2} \)

\( \begin{aligned}I &=\oint{\frac{-1f(z)dz}{(z-1)}+\frac{f(z)dz}{2(z-2)}+\frac{f(z)dz}{2(z-2)^2}} \\ I &=I_1+I_2+I_3 \end{aligned} \)

 

Applying Cauchy's integral formulae

\( \begin {aligned} \oint \frac{f(z)dz}{z-a} &=2 \pi if(a) \\ \oint \frac{f(z)dz}{(z-a)^2} &=2 \pi if'(a) \\ \end{aligned} \space \)

\( \begin {aligned} I_1 &=\oint \frac{-1f(z)dz}{(z-1)} \\ &=-2 \pi i[sin \pi + cos \pi ]\\ &=2 \pi i \end{aligned} \)

\( \begin {aligned} I_2 &=\oint \frac{f(z)dz}{2(z-2)} \\ &=2 \times (2\pi i[sin 4\pi + cos4\pi ])\\ &=4 \pi i \end{aligned} \)

\( \begin {aligned} I_3 &=\oint \frac{f(z)dz}{2(z-2)^2} \\ &=2 \times {2\pi i[4\pi (cos 4\pi - sin4\pi ) ]}\\ &=16 \pi^2 i \end{aligned} \)

Total Residue is given by,

\( \begin {aligned} I &=\oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z \\ &= 2 \pi i+4 \pi i+16 \pi^2 i \\ &= 2 \pi i(3+8\pi ) \end{aligned} \)

 

\( \begin {aligned} I &= 2 \pi i(3+8\pi ) \end{aligned} \)