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 |
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 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}
\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
\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
(-1+i)\pi | |
(1+i)\pi | |
2(1-i)\pi | |
(2+i)\pi |
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
4 | |
16 | |
4i | |
16i |
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
0 | |
2 | |
\pi i | |
2 \pi i |
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
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.
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
Bro the answer is correct. In denominator it is 2/√3. But it was written as 2√3
Q-24 upper limit pi/3 instead of x/3
Nice attempt