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GATE [EE, 2016, 1 mark] Question : The value of the integral $$\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz$$ over the contour $|z|=1$, taken in the anti-clockwise direction, would be (a) $\dfrac{24\pi i}{13}$ (b) $\dfrac{48\pi i}{13}$ (c) $\dfrac{24}{13}$ (d) $\dfrac{12}{13}$ Solution : Let $f(z)=\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}$ Poles of $f(z)$ are $\dfrac{1}{2},2+i$ and $2-i$ Only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$ $\displaystyle \text{Residue, }R=\lim_{z\to 1/2}\left[\left(z-\dfrac{1}{2}\right)f(z)\right]$ $\displaystyle R=\lim_{z\to 1/2}\left[\dfrac{2z+5}{z^2-4z+5}\right]$ $\displaystyle R=\dfrac{2\left(\dfrac{1}{2}\right)+5}{\left(\dfrac{1}{2}\right)^2-4\left(\dfrac{1}{2}\right)+5}$ $\displaystyle R=\dfrac{24}{13}$ $\therefore$ by Residue theorem, we get $\displaystyle\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz=2\pi iR=\dfrac{48\pi i}{13}$ CORRECT ANSWER : B

GATE [EC, 2015, 1 mark] Question : Let $z=x+iy$ be a complex variable. Consider that contour integartion is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE ? (a) The residue of $\dfrac{z}{z^2-1}$ at $z=1$ is $1/2$ (b) $\displaystyle\oint\limits_{C}z^2dz=0$ (c) $\displaystyle\dfrac{1}{2\pi i}\oint\limits_{C}\dfrac{1}{z}dz=1$ (d) $\bar{z}$ (complex conjugate of $z$) is analytical function Solution : $\rightarrow$ Residue of $\dfrac{z}{z^2-1}$ at $\displaystyle z=1=\lim_{z\to 1}(z-1)\dfrac{z}{z^2-1}$ $\displaystyle =\lim_{z\to 0}\dfrac{z}{z+1}$ $=\dfrac{1}{2}$ Option (A) is TRUE $\rightarrow$ $z^2$ is analytical at every point within and on circle $|z|=1$ by Cauchy's integral theorem, we get $\displaystyle\oint\limits_{C}z^2dz=0$ Option (B) is TRUE $\rightarrow\dfrac{1}{z}$ has a pole at $z=0$ $\therefore$ by Cauchy's integral theorem, we get $\displaystyle\oint\limits_{C}\dfrac{1}{z}dz=2\pi i\left[...

GATE [EC, EE, IN, 2012, 1 mark] Question : Given $\displaystyle f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$. If $C$ is a counter clockwise path in the $z$-plane such that $|z+1|=1$, the value of $\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz$ is (a) $-2$ (b) $-1$ (c) $1$ (d) $2$ Solution : Given $f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$ $\implies =\dfrac{-z+1}{(z+1)(z+3)}$ $\therefore f(z)$ has poles at $-1,-3$ $-3$ lie outside the circle $|z+1|=1$ $\therefore$ by Cauchy's Integral Formula and Cauchy's Integral Theorem, we have $\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz=\dfrac{1}{2\pi j}\left[\oint\limits_{C}\dfrac{1}{z+1}dz-0\right]$ $=\dfrac{2\pi j\left(\dfrac{-z+1}{z+3}\right)_{z=-1}}{2\pi j}$ $=\dfrac{-(-1)-1}{-1+3}$ $=\dfrac{2}{2}$ $=1$ CORRECT ANSWER : C

GATE [EE, 2013, 1 mark] Question : Square roots of $-i$, where $i=\sqrt{-1}$, are (a) $i, -i$ (b) $\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ (c) $\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)$ (d) $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right),$ $\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ Solution : As we know that $-i=i^{3}$ $=e^{i3\pi/2}$ $\therefore \sqrt{-i}=e^{i3\pi/2\times 1/2}$ $=\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ Now $,-i=\dfrac{1}{i}=e^{-i\pi/2}$ $\implies (-i)^{1/2}=e^{-i\pi/2\times 1/2}$ $=e^{-i\pi/4}$ $=\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)$ CORRECT ANSWER : B

GATE [EC, 2012, 2 mark] Question : If $f(z)=C_0+C_1z^{-1}$, then $\displaystyle\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$ is given by (a) $2\pi C_1$ (b) $2\pi(1+C_0)$ (c) $2\pi jC_1$ (d) $2\pi j(1+C_0)$ Solution : Let $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$ Given $f(z)=C_0+C_1z^{-1}$ $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+C_0+C_1/z}{z}dz$ $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{z(1+C_0)+C_1}{z^2}dz$ Singularities for $I$ is $z=0$ and $z=0$ lies inside the unit circle. $\therefore$ by Cauchy's integral theorem we have, $I=2\pi j\times(\text{residue of }\dfrac{z(1+C_0)+C_1}{z^2}\text{ at } z=0)$ $=2\pi j\left\{\dfrac{1}{1!}\dfrac{d}{dz}z^2\dfrac{z(1+C_0)+C_1}{z^2}\right\}_{z=0}$ $=2\pi j\{1+C_0\}_{z=0}$ $\therefore$ Answer $=2\pi j(1+C_0)$ CORRECT ANSWER : D

GATE [EC, 2010, 2 marks] Question : The residues of a comlex function $\displaystyle X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$ at its poles are (a) $\displaystyle\dfrac{1}{2},-\dfrac{1}{2}$ and $1$ (b) $\displaystyle\dfrac{1}{2},\dfrac{1}{2}$ and $-1$ (c) $\displaystyle\dfrac{1}{2},1$ and $\displaystyle -\dfrac{3}{2}$ (d) $\displaystyle\dfrac{1}{2},-1$ and $\displaystyle\dfrac{3}{2}$ Solution : $X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$ The poles are $z=0,1,2$ Residue at $z=0$ $\displaystyle =\lim_{z\to 0}(z-0)\frac{1-2z}{z(z-1)(z-2)}$ $=\dfrac{1-2\times 0}{(0-1)(0-2)}$ $=\dfrac{1}{2}$ Residue at $z=1$ $\displaystyle =\lim_{z\to 1}(z-1)\frac{1-2z}{z(z-1)(z-2)}$ $\dfrac{1-2\times 1}{1(1-2)}$ $=1$ Residue at $z=2$ $\displaystyle =\lim_{z\to 2}(z-2)\frac{1-2z}{z(z-1)(z-2)}$ $\dfrac{1-2\times 2}{2(2-1)}$ $=-\dfrac{3}{2}$ CORRECT ANSWER : C

GATE [CE, 2009, 2 marks] Question : The value of the integral $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$ (where $C$ is a closed curve given by $|z|=1$ is (a) $-\pi i$ (b) $\displaystyle\dfrac{\pi i}{5}$ (c) $\displaystyle\dfrac{2\pi i}{5}$ (d) $\pi i$ Solution : $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$ putting $(2z-1)(z-3)=0$ so the singularities are $z=\dfrac{1}{2}$ and $z=3$ but only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$ (the closed curve) hence by Cauchy's integral theorem $\displaystyle\dfrac{1}{2}\int\limits_{C}\dfrac{\left[\dfrac{\cos (2\pi z)}{(z-3)}\right]}{\left(z-\dfrac{1}{2}\right)}dz=\dfrac{1}{2}2\pi if(\dfrac{1}{2})$ where $f(z)=\dfrac{\cos (2\pi z)}{(z-3)}$ $\therefore\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz=\dfrac{2\pi i}{5}$ CORRECT ANSWER : C