Question 1 |
The value of the contour integral
\frac{1}{2 \pi j}\oint \left (z+\frac{1}{z} \right )^2 dz
evaluated over the unit circle |z|=1 is ______
\frac{1}{2 \pi j}\oint \left (z+\frac{1}{z} \right )^2 dz
evaluated over the unit circle |z|=1 is ______
0 | |
0.0001 | |
0.0005 | |
0.0008 |
Question 1 Explanation:
\frac{1}{2 \pi j} \oint\left(z+\frac{1}{z}\right)^{2} d z \text { where } C \text { is }|z|=1
I=\frac{1}{2 \pi j} \int_{c}^{\left(z^{2}+1\right)^{2}} d z
z=0 lies inside the circle,
\begin{aligned} I &=\frac{1}{2 \pi j}\left[\frac{2 \pi j}{1 !} \frac{d}{d z}\left(z^{2}+1\right)^{2}\right]_{2=0} \\ &=\left[\frac{d}{d z}\left(z^{2}+1\right)^{2}\right]_{z=0} \\ &=\left[2\left(z^{2}+1\right) \times 2 z\right]_{z=0}=0 \end{aligned}
Question 2 |
Which one of the following functions is analytic over the entire complex plane?
ln(z) | |
e^{1/z} | |
\frac{1}{1-z} | |
cos(z) |
Question 2 Explanation:
f(z) = \cos z is analytic every where.
Question 3 |
The contour C given below is on the complex plane z=x+jy , where j=\sqrt{-1} .
The value of the integral \frac{1}{\pi j}\oint_{C}^{}\frac{dz}{z^{2}-1} is _______.
The value of the integral \frac{1}{\pi j}\oint_{C}^{}\frac{dz}{z^{2}-1} is _______.
1 | |
2 | |
3 | |
4 |
Question 3 Explanation:
\begin{array}{l} \frac{1}{\pi \dot{i}} \oint \frac{d z}{z^{2}-1}=2\left[\frac{1}{2 \pi i} \oint_{C_{1}} \frac{d z}{(z+1)(z-1)}+\frac{1}{2 \pi i} \oint_{C_{2}} \frac{d z}{(z+1)(z-1)}\right] \\ =2\left[-\left.\left(\frac{1}{z-1}\right)\right|_{z=-1}+\left.\left(\frac{1}{z+1}\right)\right|_{z=1}\right] \\ =2\left[-\left(-\frac{1}{2}\right)+\left(\frac{1}{2}\right)\right]=2 \end{array}
Question 4 |
An integral I over a counter clock wise circle C is given by I=\oint_{0}^{c}\frac{z^{2}-1}{z^{2}+1}e^{z}dz.
If C is defined as |z| = 3, then the value of I is
If C is defined as |z| = 3, then the value of I is
-\pi i sin(1) | |
-2\pi i sin(1) | |
-3\pi i sin(1) | |
-4\pi i sin(1) |
Question 4 Explanation:
Poles are z^{2}+1=0
z=\pm i
z=i lies inside |z|=3
z=-i lies inside |z|=3
Residue at z=i is
\begin{aligned} &=\lim _{z \rightarrow i}(z-i) \frac{z^{2}-1}{(z-i)(z+i)} e^{z} \\ &=\frac{-1-1}{2 i} e^{i}=i e^{i} \end{aligned}
Residue at z=-i is
\begin{aligned} &=\lim _{z \rightarrow-i}(z+i) \frac{z^{2}-1}{(z-i)(z+i)} e^{z}\\ &=\frac{-1-1}{-2 i} e^{-i}=\frac{1}{i} e^{-i}=-i e^{-i} \end{aligned}
By Residues theorem
\begin{aligned} I &=2 \pi i\left(i e^{i}-i e^{-i}\right) \\ &=-2 \pi\left(e^{i}-e^{-i}\right) \\ &=-2 \pi(\cos 1+i \sin 1-\cos 1+i \sin 1) \\ &=-2 \pi(2 i \sin 1)=-4 \pi i \sin (1) \end{aligned}
z=\pm i
z=i lies inside |z|=3
z=-i lies inside |z|=3
Residue at z=i is
\begin{aligned} &=\lim _{z \rightarrow i}(z-i) \frac{z^{2}-1}{(z-i)(z+i)} e^{z} \\ &=\frac{-1-1}{2 i} e^{i}=i e^{i} \end{aligned}
Residue at z=-i is
\begin{aligned} &=\lim _{z \rightarrow-i}(z+i) \frac{z^{2}-1}{(z-i)(z+i)} e^{z}\\ &=\frac{-1-1}{-2 i} e^{-i}=\frac{1}{i} e^{-i}=-i e^{-i} \end{aligned}
By Residues theorem
\begin{aligned} I &=2 \pi i\left(i e^{i}-i e^{-i}\right) \\ &=-2 \pi\left(e^{i}-e^{-i}\right) \\ &=-2 \pi(\cos 1+i \sin 1-\cos 1+i \sin 1) \\ &=-2 \pi(2 i \sin 1)=-4 \pi i \sin (1) \end{aligned}
Question 5 |
The residues of a function f(z)=\frac{1}{(z-4)(z+1)} are
\frac{-1}{27} and \frac{-1}{125} | |
\frac{1}{125} and \frac{-1}{125} | |
\frac{-1}{27} and \frac{1}{5} | |
\frac{1}{125} and \frac{-1}{5} |
Question 5 Explanation:
Residue at, z=4 is
=\lim _{x \rightarrow 1}(z-4) \frac{1}{(z-4)(z+1)^{3}}=\frac{1}{(4+1)^{3}}=\frac{1}{125}
Residue at z=-1 is
=\lim _{z \rightarrow-1} \frac{1}{2 !} \frac{d^{2}}{d z^{2}}\left((z-1)^{3} \frac{1}{(z-4)(z+1)^{3}}\right)
=\lim _{z \rightarrow-1} \frac{1}{2 !}\left(\frac{2}{(z-4)^{3}}\right)=\frac{1}{(-1-4)^{3}}=\frac{-1}{125}
=\lim _{x \rightarrow 1}(z-4) \frac{1}{(z-4)(z+1)^{3}}=\frac{1}{(4+1)^{3}}=\frac{1}{125}
Residue at z=-1 is
=\lim _{z \rightarrow-1} \frac{1}{2 !} \frac{d^{2}}{d z^{2}}\left((z-1)^{3} \frac{1}{(z-4)(z+1)^{3}}\right)
=\lim _{z \rightarrow-1} \frac{1}{2 !}\left(\frac{2}{(z-4)^{3}}\right)=\frac{1}{(-1-4)^{3}}=\frac{-1}{125}
Question 6 |
The values of the integral \frac{1}{2\pi j}\oint_c \frac{e^{z}}{z-2}dz along a closed contour c in anti-clockwise direction for
(i) the point z_{0} = 2 inside the contour c, and
(ii) the point z_{0} = 2 outside the contour c,
respectively, are
(i) the point z_{0} = 2 inside the contour c, and
(ii) the point z_{0} = 2 outside the contour c,
respectively, are
(i) 2.72, (ii) 0 | |
(i) 7.39, (ii) 0 | |
(i) 0, (ii) 2.72 | |
(i) 0, (ii) 7.39 |
Question 6 Explanation:
\begin{aligned} (i) \quad\quad Z_{0}&=2 -lies inside C \\ \text { So, } \text{Res}(z)&=\lim _{z \rightarrow 2}(z-2) \cdot \frac{e^{z}}{z-2} \\ &=e^{2}=7.39 \\ \frac{1}{2 \pi i} \int \frac{e^{z}}{z-2} d z &=2 \pi i \cdot \frac{1}{2 \pi i}(7.39)=7.39 \\ (ii)\quad \quad Z_{0}&=-2\text{ lies out side C then} \\ \text{Res}(z)&=0 \\ \text{So, }\int_{c} \frac{e^{z}}{z-2} d z&=2 \pi i \frac{1}{2 \pi i}(0)=0 \end{aligned}
Question 7 |
For f(z)=\frac{sin(z)}{z^{2}}, the residue of the pole at z=0 is __________
0.5 | |
1 | |
2 | |
3 |
Question 7 Explanation:
Residue of \frac{\sin z}{z^{2}}
= coefficient of \frac{1}{2} in \left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}
= coefficient of \frac{1}{z} in \left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}
=1
= coefficient of \frac{1}{2} in \left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}
= coefficient of \frac{1}{z} in \left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}
=1
Question 8 |
Consider the complex valued function f(z)=2z^{3}+b|z|^{3} where z is a complex variable. The value of b for which the function f(z) is analytic is ________
1 | |
2 | |
3 | |
0 |
Question 8 Explanation:
f(z)=2 z^{3}+b_{1}|z|^{3}
Given that f(z) is analytic.
which is possible only when b=0
since \left|z^{3}\right| is differentiable at the origin but not analytic.
2 z^{3} is analytic everywhere
\begin{aligned} \therefore \quad f(z)&=2 z^{3}+b\left|z^{3}\right| \text{ is analytic}\\ \text{only when }\quad b&=0 \end{aligned}
Given that f(z) is analytic.
which is possible only when b=0
since \left|z^{3}\right| is differentiable at the origin but not analytic.
2 z^{3} is analytic everywhere
\begin{aligned} \therefore \quad f(z)&=2 z^{3}+b\left|z^{3}\right| \text{ is analytic}\\ \text{only when }\quad b&=0 \end{aligned}
Question 9 |
In the following integral, the contour C encloses the points 2\pi j and -2\pi j
-\frac{1}{2\pi }\oint_{0}^{C}\frac{sin z}{(z-2\pi j)^{3}}dz
The value of the integral is ________
-\frac{1}{2\pi }\oint_{0}^{C}\frac{sin z}{(z-2\pi j)^{3}}dz
The value of the integral is ________
-210.35 | |
-160.25 | |
-133.87 | |
-115.85 |
Question 9 Explanation:
\begin{aligned} I &=-\frac{1}{2 \pi} \int_{C} \frac{\sin z}{(z-2 \pi j)^{3}} d z \\ &=-\frac{1}{2 \pi} \times \frac{2 \pi j(2 \pi j)}{2 !} \\ f(z) &=\sin z \\ f^{\prime }(z) &=\cos z \\ f^{\prime \prime}(z) &=-\sin z \\ I&=-\frac{1}{2 \pi} \times 2 \pi j \frac{-\sin (2 \pi j)}{2} \\ &=\frac{1}{2} \sinh 2 \pi=-133.87 \end{aligned}
Question 10 |
If C is a circle of radius r with centre z_{0}, in the complex z-plane and if n is a non-zero integer, then \oint_{C}\frac{dz}{(z-z_{0})^{n+1}} equals
2\pi nj | |
0 | |
\frac{nj}{2\pi } | |
2\pi n |
Question 10 Explanation:
By Cauchy integral formula
\begin{array}{l} \oint \frac{f(z)}{\left(z-z_{0}\right)^{n+1} d z}=\frac{2 \pi i f^{*}\left(z_{0}\right)}{n !} \\ \oint \frac{d z}{\left(z-z_{0}\right)^{n+1}}=\frac{2 \pi i}{n !} \cdot 0=0 \end{array}
\begin{array}{l} \oint \frac{f(z)}{\left(z-z_{0}\right)^{n+1} d z}=\frac{2 \pi i f^{*}\left(z_{0}\right)}{n !} \\ \oint \frac{d z}{\left(z-z_{0}\right)^{n+1}}=\frac{2 \pi i}{n !} \cdot 0=0 \end{array}
There are 10 questions to complete.