Complex Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Complex Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 11, 2025
Latest Complex Analysis MCQ Objective Questions
Top Complex Analysis MCQ Objective Questions
Complex Analysis Question 1:
If w = u(x, y) + iv(x, y) is an analytic function of z = x + iy, then \(\rm \frac{dw}{dz}\) equals
Answer (Detailed Solution Below)
Complex Analysis Question 1 Detailed Solution
Given :- \(w=u(x,y)+iv(x,y)\) is an analytic function of \(z=x+iy\).
Concept used :- If w is an analytic function then \(\dfrac{dw}{d\overline{z} }=0\)
Solution :- As we know that
\(\dfrac{\partial w}{\partial z}=\dfrac{1}{2}(\dfrac{\partial w}{\partial x}-i\dfrac{\partial w}{\partial y})\) and \(\dfrac{\partial w}{\partial \bar{z}}=\dfrac{1}{2}(\dfrac{\partial w}{\partial x}+i\dfrac{\partial w}{\partial y})\)
subtracting both equation and putting \(\dfrac{dw}{d\overline{z} }=0\), we get
\(\dfrac{dw}{dz}=-i\dfrac{\partial w}{\partial y}\)
Complex Analysis Question 2:
Let f : ℂ → ℂ be a real-differentiable function. Define u, v : ℝ2 → ℝ by u(x, y) = Re f(x + i y) and v(x, y) = Im f(x + iy), x, y ∈ ℝ.
Let ∇u = (ux, uy) denote the gradient. Which one of the following is necessarily true?
Answer (Detailed Solution Below)
Complex Analysis Question 2 Detailed Solution
Concept:
(i) A complex function f(z) is entire function if it is analytic in whole complex plane.
(ii) If a complex function f(z) = u + iv is entire then it satisfy C-R equation i.e., ux = vy, uy = - vx
Explanation:
f : ℂ → ℂ is a real-differentiable function.
u, v : ℝ2 → ℝ by u(x, y) = Re f(x + i y) and v(x, y) = Im f(x + iy), x, y ∈ ℝ.
Also, ∇u = (ux, uy)
(1): Then "For c1, c2 ∈ ℂ, the level curves u = c1 and v = c2 are orthogonal wherever they intersect" this statement will satisfy only if f(z) is analytic function.
(1) is false
(3): f(z) is entire function so ux = vy, uy = - vx
then ∇u . ∇v = (ux, uy) . (vx, vy) = uxvx + uyvy = uxvx - vxux = 0 at every point.
(3) is true and (2) is false
(4): ∇u . ∇v = 0
⇒ (ux, uy) . (vx, vy) = 0
⇒ uxvx + uyvy = 0
⇒ uxvx = - uyvy
which does not imply ux = vy, uy = - vx
f is not an entire function.
(4) is false
Complex Analysis Question 3:
\(\displaystyle \int_{|z+1|=2} \frac{z^2}{4-z^2} d z=\)
Answer (Detailed Solution Below)
Complex Analysis Question 3 Detailed Solution
Given -
\(\displaystyle \int_{|z+1|=2} \frac{z^2}{4-z^2} d z=\)
Concept -
If singular point z = c of f(z) lies in | z - a | = r then \(\displaystyle \int_{|z-a|=r} F(z) dz = 2\pi i \times Rez(f(z))_{ z = c}\)
If singular point z = c of f(z) does not lie in | z - a | = r then \(\displaystyle \int_{|z-a|=r} F(z) dz = 2\pi i \times Rez(f(z))_{ z = c}=0\)
Explanation -
\(\displaystyle \int_{|z+1|=2} \frac{z^2}{4-z^2} d z=\displaystyle \int_{|z+1|=2} F(z) dz\)
Where \(f(z) = \frac{z^2}{4-z^2}\)
For Singularity - \(4-z^2=0\)
⇒ \(z = -2, +2\)
F(z) has singularity at z = 2 and z = -2 But the singularity z = 2 does not lie in \(|z+1|=2\) Hence the integral should be zero for z = 2.
Now singularity z = -2 lies in \(|z+1|=2\) Hence we have to calculate the integral using the above concept -
So, \(\displaystyle \int_{|z+1|=2} F(z) dz = 2\pi i \times Rez(f(z)) \ \ at \ \ z = -2\) .........(i)
⇒ \(Rez(f(z))|_{z=-2} = lim_{z \to -2} (z+2) \frac{z^2}{4-z^2}\)
= \( lim_{z \to -2} \frac{z^2}{2-z}=\frac{4}{4}=1\)
Put this value in the above equation we get -
⇒ \(\displaystyle \int_{|z+1|=2} F(z) dz = 2\pi i \times Rez(f(z))_{ z = -2}= 2 \pi i\)
Hence the option (iii) is correct.
Complex Analysis Question 4:
Let f be an entire function that satisfies |f(z)| ≤ ey for all z = x + iy ∈ ℂ, where x, y ∈ ℝ. Which of the following statements is true?
Answer (Detailed Solution Below)
Complex Analysis Question 4 Detailed Solution
Explanation:
f be an entire function that satisfies |f(z)| ≤ ey for all z = x + iy ∈ ℂ
(1): f(z) = ce−iz
So |f(z)| = |ce−iz| = |ce-i(x + iy)| = |ce-ix ey| ≤ |c|ey ≤ ey for some c ∈ ℂ with |c| ≤ 1 (as |e-ix| ≤ 1 for all x ∈ \(\mathbb R\))
Option (1) is correct
(2): f(z) = ceiz
So |f(z)| = |ceiz| = |cei(x + iy)| = |ceix e-y| ≤ e-y for some c ∈ ℂ with |c| ≤ 1 (as |e-ix| ≤ 1 for all x ∈ \(\mathbb R\))
Option (2) is false
(3): f(z) = e−ciz
So |f(z)| = |e−ciz | = |e-ci(x + iy)| = |e-cix ecy| ≤ ecy ≤ ey for c = 1 only (as |e-ix| ≤ 1 for all x ∈ \(\mathbb R\))
Option (3) is false
(4): f(z) = eciz
So |f(z)| = |eciz | = |eci(x + iy)| = |ecix e-cy| ≤ e-cy ≤ ey for c = - 1 only (as |e-ix| ≤ 1 for all x ∈ \(\mathbb R\))
Option (4) is false
Complex Analysis Question 5:
If R is the radius of convergence of any power series then what is the interval of convergence?
Answer (Detailed Solution Below)
Complex Analysis Question 5 Detailed Solution
Explanation:
If R is the radius of convergence of \(\sum a_nx^n\) then the series converges when |x| < R
i.e., when - R < x < R
So interval of convergence is (- R, R)
(1) is correct
Complex Analysis Question 6:
How many roots does the polynomial z100 - 50z30 + 40z10 + 6z + 1 have in the open disc {z ∈ ℂ : |z| < 1}?
Answer (Detailed Solution Below)
Complex Analysis Question 6 Detailed Solution
Concept:
Rouche’s Theorem: If f(z) and g(z) are two analytic functions within and on a simple closed curve C such that |f(z)| < |g(z)| at each point on C, then both f(z) + g(z) and g(z) have the same number of roots inside C.
Explanation:
z100 - 50z30 + 40z10 + 6z + 1 and the open disc {z ∈ ℂ : |z| < 1}
Let f(z) = z100 + 40z10 + 6z + 1 and g(z) = - 50z30
Then |f(z)| = |z100 + 40z10 + 6z + 1| ≤ |z100|+ 40|z10| + 6|z| + 1 < 1 + 40 + 6 + 1 = 48
and |g(z)| = | - 50z30| = 50|z30| < 50
Hnece |f(z)| < |g(z)| inside {z ∈ ℂ : |z| < 1}
Then By Rouche's theorem,
f(z)+g(z) and g(z) has same roots inside {z ∈ ℂ : |z| < 1}
Now, g(z) = - 50z30 has 30 roots inside {z ∈ ℂ : |z| < 1}
Therefore z100 - 50z30 + 40z10 + 6z + 1 has 30 roots inside {z ∈ ℂ : |z| < 1}
(3) is correct
Complex Analysis Question 7:
The radius of convergence of the series \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}z^{2n}}{n!}\) is
Answer (Detailed Solution Below)
Complex Analysis Question 7 Detailed Solution
Concept:
The radius of convergence of the series \(\sum_{n=1}^{\infty}a_nz^{kn}\) is \(\frac{1}{R}=\left(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\right)^{1/k}\)
Explanation:
Here in the series \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}z^{2n}}{n!}\), \(a_n=\frac{(-1)^{n-1}}{n!}\) and k = 2
The radius of convergence of the series is
1/R = \(\left(\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n-1}}{(n+1)!}}{\frac{(-1)^{n-1}}{(n)!}}\right|\right)^{1/2}\)
⇒ 1/R = \(\left(\lim_{n\to\infty}\frac{n!}{(n+1)!}\right)^{1/2}\)
⇒ 1/R = \(\left(\lim_{n\to\infty}\frac{1}{(n+1)}\right)^{1/2}\)
⇒ \(R= \infty\)
(2) is correct
Complex Analysis Question 8:
Which of the following is the Taylor series expansion of f(z) = \(\frac{(z^2-1)}{(z+2)(z+3)}\) in |z| < 2
Answer (Detailed Solution Below)
Complex Analysis Question 8 Detailed Solution
Explanation:
f(z) = \(\frac{(z^2-1)}{(z+2)(z+3)}\) = 1 - \(\frac{(5z+7)}{(z+2)(z+3)}\)
⇒ f(z) = 1 - (\(\frac8{z+3}-\frac{3}{z+2}\))
⇒ f(z) = 1 + \(\frac{3}{z+2}-\frac8{z+3}\)
Now for |z| < 2 ⇒ \(|\frac{z}{2}|<1\) and \(|\frac{z}{3}|<1\)
⇒ f(z) = 1 + \(\frac{3}{2(1+\frac z2)}-\frac8{3(1+\frac z3)}\)
⇒ f(z) = 1 + \(\frac32(1+\frac z2)^{-1}\) - \(\frac83(1+\frac z3)^{-1}\)
⇒ f(z) = 1 + \(\frac32(1-\frac z2+\frac{z^2}{4}-...)\) - \(\frac83(1-\frac z3+\frac{z^2}{9}-...)\) (\((1+x)^{-1}=1-x+x^2-x^3+x^4-...\) for |x| < 1)
(1) is correct
Complex Analysis Question 9:
Let R denote the radius of convergence of power series \(\rm \displaystyle \sum_{k=1}^{\infty} k x^{k}\). Then
Answer (Detailed Solution Below)
Complex Analysis Question 9 Detailed Solution
Given -
Let R denote the radius of convergence of power series \(\rm \displaystyle \sum_{k=1}^{\infty} k x^{k}\).
Concept -
If the power series is given by \(\sum_{n=0}^{\infty} a_n x^{n}\)
then Radius of convergence R of the power series is
\(\frac{1}{R}= lim_{n \to \infty }\frac{a_{n+1}}{a_n}\)
And The interval of convergence for the power series is \(|x|
Explanation -
\(\frac{1}{R}= lim_{k \to \infty }\frac{{k+1}}{k}\)
⇒ R = 1
Hence option (iv) is false.
Now the interval of convergence for the power series is \(|x|<1\)
Now we check the convergence of the power series at end points -
Now at x = 1
put this value in the given series we get the series -
\(\rm \displaystyle \sum_{k=1}^{\infty} k \)
Clearly the series is divergent.
Now at x = -1
put this value in the given series we get the series -
\(\rm \displaystyle \sum_{k=1}^{\infty} (-1)^kk \)
Clearly the series is also divergent.
Hence the power series is convergent at (-1,1) or (-R,R)
Hence option (iii) is true.
Complex Analysis Question 10:
Let g(z) = z3 and and f(z) = z3 - z - 1. Then the value of \({1\over 2\pi i}\int_Cg(z){f'(z)\over f(z)}dz\) where C contains all the zeros of f(z) is
Answer (Detailed Solution Below)
Complex Analysis Question 10 Detailed Solution
Concept:
Argument theorem: Let f be meromorphic function and C be simple closed contour such that no zero or pole of f lies inside C. Let a1, a2,…,ak be zeros of f of order n1, n2,…,nk respectively and f(z) has no pole in C. Then
\({1\over 2\pi i}\int_Cg(z){f'(z)\over f(z)}dz\) = \(\sum_{r=1}^kn_rg(a_r)\)
Explanation:
g(z) = z3 and and f(z) = z3 - z - 1.
Let a, b, c are the zeros of f(z) lies in C then
a + b + c = 0
a2 + b2 + c2 = -1 and
abc = 1
Now,
\({1\over 2\pi i}\int_Cg(z){f'(z)\over f(z)}dz\) = a3 + b3 + c3
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca) + 3abc
= 0(-1) + 3 = 3
Option (1) is true.