# Complex Variables

 Question 1
The value of $k$ that makes the complex-valued function

$f(z)=e^{-kx}(\cos 2y -i \sin 2y)$

analytic, where $z=x+iy$, is _________. (Answer in integer)
 A 1 B 2 C 3 D 4
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Question 1 Explanation:
\begin{aligned} & f(z)=e^{-k x} \cos 2 y-i e^{-k x} \sin 2 y \\ & \text { Suppose }=e^{-k x} \cos 2 y=4(x, y) \\ & =-i e^{-k x} \sin 2 y=v(x, y) \end{aligned}
If function is analytical then it satisfy the equation
\begin{aligned} & \frac{\partial u}{\partial x}=-k e^{-k x} \cos 2 y &...(i)\\ & \frac{\partial u}{\partial y}=-2 e^{-k x} \sin 2 y &...(ii)\\ & \frac{\partial v}{\partial x}=-k e^{-k x} \sin 2 y &...(iii)\\ & \frac{\partial v}{\partial y}=-2 e^{-k x} \cos 2 y &...(iv) \end{aligned}
Cauchy-Riemann equation $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
by putting equation (i), (iv) and solve then $k=2$
 Question 2
Given $z=x+iy,i=\sqrt{-1}$ is a circle of radius 2 with the centre at the origin. If the contour C is traversed anticlockwise, then the value of the integral $\frac{1}{2}\int_{c}^{}\frac{1}{(z-i)(z+4i)}dz$ is _______(round off to one decimal place).
 A 0.2 B 0.4 C 0.6 D 0.8
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Question 2 Explanation:
Given contour is a circle at centre (0,0) and radius 2 given function is $\frac{1}{(z-i)(z+4i)}$ here $z = i$ is a singular point lies now by caucus integral formula

\begin{aligned} &\int \frac{1}{(z-i)(z+4i)}dz=\int \frac{\left ( \frac{1}{z+4} \right )}{z-i}dz=2 \pi i \times f(i)\\ &f(z)=\frac{1}{z+4i}\\ &f(i)=\frac{1}{5i}\\ &2 \pi i \times f(i)=2 \pi i \times \frac{1}{5i}=\frac{2\pi}{5}\\ &\text{Now }\frac{1}{2 \pi}\int_{c}\frac{1}{(z-1)(z+4i)}dz=\frac{1}{2 \pi}+\frac{2 \pi}{5}=\frac{1}{5} \end{aligned}

 Question 3
The value of the integral
$\oint \left (\frac{6z}{2z^4-3z^3+7z^2-3z+5} \right )dz$
evaluated over a counter-clockwise circular contour in the complex plane enclosing only the pole $z = i$, where $i$ is the imaginary unit, is
 A $(-1+i)\pi$ B $(1+i)\pi$ C $2(1-i)\pi$ D $(2+i)\pi$
GATE ME 2022 SET-1   Engineering Mathematics
Question 3 Explanation:
\begin{aligned} f(z)&=\frac{6z}{(2z^4-3\tau ^3+7\tau ^2-3x+5)}\\ &=\frac{6z}{(2z^4+2z^2)(5z^2+5)-3z^3-3z}\\ &=\frac{6z}{qz^2(z^2+1)+5(z^2+1)-3z(z^2+1)}\\ &=\frac{6z}{(z^2+1)(2z^2-3z+5)}\\ \therefore \oint f(z)dz&=2 \pi i \lim_{z \to 1}(z-i)f(z)\\ &=2\pi i \lim_{z \to i}\frac{6z}{(z+i)(2z^2-3z+5)}\\ &=2 \pi i \frac{6i}{2i(3-3i)}\\ &=\frac{2 \pi i}{(1-i)}\\ &=\pi(i-1) \end{aligned}
 Question 4
Value of $(1+i)^8$, where $i=\sqrt{-1}$, is equal to
 A 4 B 16 C 4i D 16i
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Question 4 Explanation:
\begin{aligned} (1+i)^{8} & \\ \mathrm{z} &=1+i \mathrm{r}=|z|=\sqrt{2} \\ \theta &=\frac{\pi}{4} \\ (1+i)^{8} &=\left(\sqrt{2} e^{i \frac{\pi}{4}}\right)^{8} \\ &=16 \times e^{i \times 2 \pi} \\ &=16(\cos 2 \pi+i \sin 2 \pi) \\ &=16 \times 1=16 \end{aligned}
 Question 5
Let C represent the unit circle centered at origin in the complex plane, and complex variable, $z=x+iy$. The value of the contour integral $\oint _C\frac{\cosh 3z}{2z}dz$ (where integration is taken counter clockwise) is
 A 0 B 2 C $\pi i$ D $2 \pi i$
GATE ME 2021 SET-1   Engineering Mathematics
Question 5 Explanation:
Pole of $f(c)$ is $z=0$ simple pole.
Reduce at z=0
$R_{b} f(z)=\lim _{z \rightarrow 0}(z-0) f(z)=\lim _{z \rightarrow 0} \frac{\cosh (3 z)}{2}=\lim _{z \rightarrow 0} \frac{e^{3 z}+e^{-3 z}}{2 \times 2}=\frac{1}{2}$
By Cauchy Riemann theorem,
$I=2 \pi\left(\frac{1}{2}\right)=\pi i$

There are 5 questions to complete.

### 4 thoughts on “Complex Variables”

1. que no 14, ans should be option B) 30
value of arg Z2 should be 30 so the final ans should be 60-30=30