← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q3c — Step-by-Step Solution

15 marks · Section A

Cauchy's residue theorem · Complex Analysis · asked 5× in 14 yrs · Read the full method →

Question

Let $\gamma:[0,1]\to\mathbb C$ be the curve $\gamma(t)=e^{2\pi it},\ 0\le t\le1$. Find, giving justifications, the value of the contour integral $\displaystyle\int_\gamma\frac{dz}{4z^2-1}$.

Technique

Residue theorem on the unit circle; factor $4z^2-1=(2z-1)(2z+1)$; simple-pole residues via $g/h'=1/(8z)$; sum the (interior) residues.

Solution

$\gamma$ is the unit circle $|z|=1$ traversed once counterclockwise. We locate the poles of the integrand relative to $\gamma$, compute residues at those inside, and apply the Cauchy residue theorem.

Step 1 — Identify the contour and the singularities

$\gamma(t)=e^{2\pi i t}$, $t\in[0,1]$, is the positively oriented unit circle $|z|=1$. Factor the denominator:

so the integrand $f(z)=\dfrac1{4z^2-1}$ has simple poles at

Both satisfy $|z|=\tfrac12<1$, so both poles lie inside $\gamma$. (Neither lies on $\gamma$, so the integral is well defined.)

Step 2 — Residues at the simple poles

For a simple pole, $\operatorname{Res}_{z_0}\dfrac{g(z)}{h(z)}=\dfrac{g(z_0)}{h'(z_0)}$ with $g=1,\ h=4z^2-1,\ h'=8z$.

(Equivalently via partial fractions $\dfrac1{4z^2-1}=\dfrac{1/4}{z-1/2}-\dfrac{1/4}{z+1/2}$, the residues are the coefficients $\tfrac14$ and $-\tfrac14$.)

Step 3 — Apply the residue theorem

Since $f$ is analytic inside and on $\gamma$ except at the two interior simple poles, the residue theorem gives

Therefore

Answer