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)
f(z)=e^{-kx}(\cos 2y -i \sin 2y)
analytic, where z=x+iy, is _________. (Answer in integer)
1 | |
2 | |
3 | |
4 |
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
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 |
Let w^{4}=16 j. Which of the following cannot be a value of w ?
2 e^{\frac{j 2 \pi}{8}} | |
2 e^{\frac{j \pi}{8}} | |
2 e^{\frac{j 5 \pi}{8}} | |
2 e^{\frac{j 9 \pi}{8}} |
Question 2 Explanation:
w=(2) j^{1 / 4}
w=2(0+j)^{1 / 4}
w=2\left[e^{j(2 n+1) \pi / 2}\right]^{1 / 4} =2\left[e^{j(2 n+1) \pi / 8}\right]
For n=0, w=e^{j \pi / 8}
For n=2, w=2 e^{5 \pi j / 8}
For n=4, w=2 e^{9 \pi j / 8}
w=2(0+j)^{1 / 4}
w=2\left[e^{j(2 n+1) \pi / 2}\right]^{1 / 4} =2\left[e^{j(2 n+1) \pi / 8}\right]
For n=0, w=e^{j \pi / 8}
For n=2, w=2 e^{5 \pi j / 8}
For n=4, w=2 e^{9 \pi j / 8}
Question 3 |
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).
0.2 | |
0.4 | |
0.6 | |
0.8 |
Question 3 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}
\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 4 |
Let R be a region in the first quadrant of the xy plane enclosed by a closed curve C
considered in counter-clockwise direction. Which of the following expressions does
not represent the area of the region R?
\int \int _R dxdy | |
\oint _c xdy | |
\oint _c ydx | |
\frac{1}{2}\oint _c( xdy-ydx) |
Question 4 Explanation:
Using green theorem?s
\oint _cF_1dx+F_2dy=\int \int _R\left ( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\right )dxdy
Check all the options:
\oint xdy=\int \int _R\left ( \frac{\partial x}{\partial x} -0\right )dxdy=\int \int _Rdxdy
\frac{1}{2}\oint xdy-ydx=\frac{1}{2}\int \int _R(1+1)dxdy=\int \int _Rdxdy
\oint ydx=\int \int _R(-1)dxdy=-\int \int _Rdxdy
Hence, \oint ydx is not represent the area of the region.
\oint _cF_1dx+F_2dy=\int \int _R\left ( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\right )dxdy
Check all the options:
\oint xdy=\int \int _R\left ( \frac{\partial x}{\partial x} -0\right )dxdy=\int \int _Rdxdy
\frac{1}{2}\oint xdy-ydx=\frac{1}{2}\int \int _R(1+1)dxdy=\int \int _Rdxdy
\oint ydx=\int \int _R(-1)dxdy=-\int \int _Rdxdy
Hence, \oint ydx is not represent the area of the region.
Question 5 |
Consider the following series:
\sum_{n=1}^{\infty }\frac{n^d}{c^n}
For which of the following combinations of c,d values does this series converge?
\sum_{n=1}^{\infty }\frac{n^d}{c^n}
For which of the following combinations of c,d values does this series converge?
c=1,d=-1 | |
c=2,d=1 | |
c=0.5,d=-10 | |
c=1,d=-2 |
Question 5 Explanation:
\begin{aligned}
\Sigma u_n&=\Sigma \frac{n}{2^n}\\
&\text{Ratio test; }\\
\lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{n+1}{2^{n+1}} \times \frac{2^n}{n}=\frac{1}{2}\\
\frac{1}{2}& \lt 2
\end{aligned}
\therefore By ratio test, \Sigma u_n is convergent.
(A) c=1, d=-1
\Sigma u_n=\Sigma \frac{1}{n} is divergent by P-test
(B) c=0.5, d=-10
\begin{aligned} \Sigma u_n&=\Sigma \frac{n^{-10}}{(0.5)^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{(n+1)^{-10}}{(0.5)^{n+1}} \times \frac{(0.5)^n}{n^{10}}\\ &=\frac{1}{0.5}=2\\ 2& \gt 1 \end{aligned}
\therefore By ratio test, \Sigma u_n is divergent.
(D) c=1,d=-2 \Sigma u_n=\Sigma \frac{n^{-2}}{(1)^n}=\Sigma \frac{1}{n^2}
\Sigma u_n is convergent by P-test.
\therefore By ratio test, \Sigma u_n is convergent.
(A) c=1, d=-1
\Sigma u_n=\Sigma \frac{1}{n} is divergent by P-test
(B) c=0.5, d=-10
\begin{aligned} \Sigma u_n&=\Sigma \frac{n^{-10}}{(0.5)^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{(n+1)^{-10}}{(0.5)^{n+1}} \times \frac{(0.5)^n}{n^{10}}\\ &=\frac{1}{0.5}=2\\ 2& \gt 1 \end{aligned}
\therefore By ratio test, \Sigma u_n is divergent.
(D) c=1,d=-2 \Sigma u_n=\Sigma \frac{n^{-2}}{(1)^n}=\Sigma \frac{1}{n^2}
\Sigma u_n is convergent by P-test.
There are 5 questions to complete.
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Que no. 21 incorrect data
We have review it with original paper form official GATE website and found it correct.
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