Question 1 |
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 1 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 2 |
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 2 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 3 |
Value of (1+i)^8, where i=\sqrt{-1}, is equal to
4 | |
16 | |
4i | |
16i |
Question 3 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 4 |
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 4 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
Question 5 |
The function f(z) of complex variable z = x + iy, where i = \sqrt{-1}, is given as
f(z) = (x^3 - 3xy^2) + iv(x, y). For this function to be analytic, v(x, y) should be
(3xy^2-y^3)+constant | |
(3 x^2 y^2-y^3)+constant | |
(x^3-3x^2 y^3)+constant | |
(3x^2y-y^3)+constant |
Question 5 Explanation:
\begin{aligned} f(z) &=u+i v \\ u &=x^{3}-3 x y^{2}, \quad v=v(x, y)\\ &\text{For }f(z) \text{ to be Analytical,}\\ u_{x} &=3 x^{2}-3 y^{2}=v_{y} \\ u_{y} &=-6 x y=-v_{x} \\ v_{x} &=6 x y \text { by integrating w.r.t } x \Rightarrow v=3 x^{2} y+C_{1} \\ v_{y} &=3 x^{2}-3 y^{2} \text { by integrating w.r.t } y \Rightarrow v=3 x^{2} y-y^{3}+C_{2} \\ v &=\left(3 x^{2} y-y^{3}\right)+\text { constant }\left(C_{1}=-y^{3}\right) \end{aligned}
Question 6 |
An analytic function of a complex variable z=x+iy(i=\sqrt{-1}) is defined as
f(z)=x^2-y^2+i\psi (x,y)
where \psi (x,y) is a real function. The value of the imaginary part of f(z) at z=(1+i) is ___________ (round off to 2 decimal places)
f(z)=x^2-y^2+i\psi (x,y)
where \psi (x,y) is a real function. The value of the imaginary part of f(z) at z=(1+i) is ___________ (round off to 2 decimal places)
1 | |
2 | |
3 | |
4 |
Question 6 Explanation:
\begin{aligned} f(z) &=\phi+i \psi \text { is analytic } \\ &\text{C-R equations} \\ \phi _x&=\psi _x \\ \phi_{y}&=-\psi _y \\ \phi &=x^{2}-y^{2} \\ \phi_{x} &=2 x=\psi_{y}
\\ \phi_{y} &=-2 y=-\psi_{x} \\ \psi_{x} &=2 y \quad \Rightarrow \psi=2 x y+C_{1} \\ \psi_{y} &=2 x \quad \Rightarrow \psi=2 x y+C_{2}\\ \text{Comparing }\psi&=2 x y+C \text{ valid for all C put C=0 }\\ \psi(1+i) &\Rightarrow(x=1 \quad y=1)\\ \therefore \psi&=2 \end{aligned}
Question 7 |
Which of the following function f(z), of the complex variable z, is NOT analytic at all the points of the complex plane?
f(z)=z^2 | |
f(z)=e^z | |
f(z)=\sin z | |
f(z)=\log z |
Question 7 Explanation:
logz is not analytic at all points.
Question 8 |
An analytic function f(z) of complex variable z=x+iy may be written as f(z)=u(x,y)+iv(x,y). Then u(x,y) and v(x,y) must satisfy
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} |
Question 8 Explanation:
Given that the complex function f(z)=u(x,y)+ i v(x,y) is an analytic function.
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
Question 9 |
Let z be a complex variable. For a counter-clockwise integration around a unit circle C ,centred at origin,
\mathrm{\oint_{c} \frac{1}{5z-4}dz}=A\pi i
the value of A is
\mathrm{\oint_{c} \frac{1}{5z-4}dz}=A\pi i
the value of A is
\frac{2}{5} | |
\frac{1}{2} | |
2 | |
\frac{4}{5} |
Question 9 Explanation:
\begin{aligned} 5 z-4&=0 \\ z&=\frac{4}{5} \text{ lies inside circle,}\\ |z|&=1\\ \int \frac{1}{(5 z-4)} d z &=A \pi i \\ \frac{1}{5} \int \frac{1}{\left(z-\frac{4}{5}\right)} d z &=A \pi i \\ \int \frac{\left(\frac{1}{5}\right)}{\left(z-\frac{4}{5}\right)} d z &=2 \pi i \cdot f\left(\frac{4}{5}\right) \\ & =2 \pi i \times\left(\frac{1}{5}\right)\\ &=\frac{2}{5} \pi i \\ A &=\frac{2}{5}=0.4 \end{aligned}
Question 10 |
F(z) is a function of the complex variable z = x + iy given by
F(z)=iz+kRe(z)+ i \; lm(z)
For what value of k will F(z) satisfy the Cauchy-Riemann equations?
F(z)=iz+kRe(z)+ i \; lm(z)
For what value of k will F(z) satisfy the Cauchy-Riemann equations?
0 | |
1 | |
-1 | |
y |
Question 10 Explanation:
\begin{aligned} F(z) &=i z+k \operatorname{Re}(z)+i \operatorname{Im}(z) \\ u+i v &=i(x+i y)+k x+i y \\ u+i v &=k x-y+i(x+y) \\ u &=k x-y, v=x+y \\ u_{x} &=k, u_{y}=-1 \\ V &=x+y \\ v_{x} &=1 \\ v_{y} &=1 \\ u_{x} &=v_{y} \\ k &=1 \end{aligned}
There are 10 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