GATE 2025 PH – Question 55
Consider $$I=\frac1{2\pi i}\oint\frac{z^4-1}{(z-a/b)(z-b/a)}\,dz,$$ on the unit circle centered at the origin, where $a,b>0$. Which option(s) is/are correct?
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Correct answer: (A) $I=5/8$ when $a=1,b=2$; (C) $I=5/8$ when $a=2,b=1$
Explanation
$I=\dfrac{1}{2\pi i}\oint\dfrac{z^4-1}{(z-a/b)(z-b/a)}dz$ over the unit circle. By the residue theorem, $I$ is the sum of the residues at the poles inside $|z|=1$.
The poles are $r=a/b$ and $1/r=b/a$. Exactly one of them lies inside the unit circle (their product is 1), the one with modulus below 1. Call it $r<1$.
**Residue at $z=r$:**
$$\text{Res}=\frac{r^4-1}{r-1/r}=\frac{(r^2-1)(r^2+1)}{(r^2-1)/r}=r\,(r^2+1).$$
**Evaluate** for each case:
- **A. $a=1,b=2$:** $r=\tfrac12$, $I=\tfrac12\left(\tfrac14+1\right)=\tfrac58$. ✓
- **B. $a=1,b=3$:** $r=\tfrac13$, $I=\tfrac13\left(\tfrac19+1\right)=\tfrac{10}{27}\neq\tfrac{10}3$. ✗
- **C. $a=2,b=1$:** the poles are 2 and $\tfrac12$, so $r=\tfrac12$, $I=\tfrac58$. ✓
- **D. $a=3,b=2$:** the poles are $\tfrac32$ and $\tfrac23$, so $r=\tfrac23$, $I=\tfrac23\left(\tfrac49+1\right)=\tfrac{26}{27}\neq\tfrac58$. ✗
Answer **A and C**.