GATE 2025 EE Q.55: solution and answer

Official answer: 0Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Start with the integrand. $f(z)=jz$ is a polynomial in $z$, so it is analytic at every point of the finite plane (an entire function). It has no singularity inside or on $|z|=1$.

2. Apply Cauchy's integral theorem: if $f$ is analytic on and inside a simple closed contour $C$, then ${\oint }_{C}f(z)dz=0$. Traversing the contour clockwise instead of anticlockwise only changes the sign of the integral, and $-0=0$.

3. Check directly by parametrisation. The clockwise unit circle is $z=e^{-j\theta }$, $0\le \theta \le 2\pi$, so $dz=-j{e}^{-j\theta }d\theta$.

4. $${\oint }_{C}jzdz={\int }_0^{2\pi }j{e}^{-j\theta }\left(-j{e}^{-j\theta }\right)d\theta ={\int }_0^{2\pi }{e}^{-2j\theta }d\theta ={\left[\frac{e^{-2j\theta }}{-2j}\right]}_0^{2\pi }=\frac{1-1}{-2j}=0$$ (using $-j\cdot j=-j^2=1$ and $e^{-4j\pi }=1$).

5. A third check: $jz$ has the antiderivative $\frac{j{z}^2}{2}$, which is single-valued in the whole plane, so its integral around any closed path is $0$.

6. Therefore ${\oint }_{C}f(z)dz=0$.