Complex Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Complex Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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., u_x = v_y, u_y = - v_x

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

Concept:

A complex function f(z) is differentiable then $\frac{\partial f}{\partial \overline{z}}$ = 0

Explanation:

f(z) = |z|2, z ∈ ℂ

so f(z) = x² + y²

then f(z) is continuous everywhere

Now, f(z) = |z|2 = $z\overline{z}$

So, $\frac{\partial f}{\partial \overline{z}}$ = z 

Therefore $\frac{\partial f}{\partial \overline{z}}$ = 0 at z = 0 only

 i.e., f(z) is differentiable at the origin only

Hence f(z) is continuous everywhere but nowhere differentiable except at the origin.

(3) is correct

Given -

 ${\displaystyle {\int }_{|z+1|=2}\frac{z^2}{4-z^2}dz=}$

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 -

  ${\displaystyle {\int }_{|z+1|=2}\frac{z^2}{4-z^2}dz={\displaystyle {\int }_{|z+1|=2}F(z)dz}}$

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, ${\displaystyle {\int }_{|z+1|=2}F(z)dz=2\pi i\times Rez(f(z))atz=-2}$  .........(i)

⇒ $Rez(f(z)){|}_{z=-2}=li{m}_{z\to -2}(z+2)\frac{z^2}{4-z^2}$

= $li{m}_{z\to -2}\frac{z^2}{2-z}=\frac{4}{4}=1$

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.

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

Explanation: 

If R is the radius of convergence of $\sum a_n{x}^n$ then the series converges when |x| < R

i.e., when - R < x < R

So interval of convergence is (- R, R)

(1) is correct

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

Explanation:

f(z) = $\frac{(z^2-1)}{(z+2)(z+3)}$ = 1 - $\frac{(5z+7)}{(z+2)(z+3)}$ 

⇒ f(z) = 1 - ($\frac{8}{z+3}-\frac{3}{z+2}$)

⇒ f(z) = 1 + $\frac{3}{z+2}-\frac{8}{z+3}$

Now for |z| < 2 ⇒ $|\frac{z}{2}|<1$ and $|\frac{z}{3}|<1$

⇒ f(z) = 1 + $\frac{3}{2(1+\frac{z}{2})}-\frac{8}{3(1+\frac{z}{3})}$

⇒ f(z) = 1 + $\frac{3}{2}(1+\frac{z}{2}{)}^{-1}$ - $\frac{8}{3}(1+\frac{z}{3}{)}^{-1}$

⇒ f(z) = 1 + $\frac{3}{2}(1-\frac{z}{2}+\frac{z^2}{4}-...)$ - $\frac{8}{3}(1-\frac{z}{3}+\frac{z^2}{9}-...)$ ($(1+x{)}^{-1}=1-x+x^2-x^3+x^4-...$ for |x| < 1) 

(1) is correct

Given -

Let R denote the radius of convergence of power series ${\displaystyle \sum _{\mathrm{k}=1}^{\mathrm{\infty }}\mathrm{k}{\mathrm{x}}^{\mathrm{k}}}$. 

Concept -

If the power series is given by $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$

 then Radius of convergence R of the power series is 

$\frac{1}{R}=li{m}_{n\to \mathrm{\infty }}\frac{a_{n+1}}{a_n}$

 And The interval of convergence for the power series is \(|x| 

Explanation -

$\frac{1}{R}=li{m}_{k\to \mathrm{\infty }}\frac{k+1}{k}$

⇒ 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 -

${\displaystyle \sum _{\mathrm{k}=1}^{\mathrm{\infty }}\mathrm{k}}$

Clearly the series is divergent.

Now at x = -1

put this value in the given series we get the series -

${\displaystyle \sum _{\mathrm{k}=1}^{\mathrm{\infty }}(-1{)}^{\mathrm{k}}\mathrm{k}}$

Clearly the series is also divergent.

Hence the power series is convergent at (-1,1) or (-R,R)

Hence option (iii) is true.

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,…,a_k be zeros of f of order n₁, n₂,…,n_k respectively and f(z) has no pole in C. Then
 $\frac{1}{2\pi i}{\int }_{C}g(z)\frac{f^{\prime }(z)}{f(z)}dz$ = $\sum _{r=1}^k{n}_{r}g(a_r)$

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, 

$\frac{1}{2\pi i}{\int }_{C}g(z)\frac{f^{\prime }(z)}{f(z)}dz$ = a³ + b³ + c³

                           = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

                           = 0(-1) + 3 = 3

Option (1) is true.

Concept:

Cauchy Residue Theorem:

Let $f(z)$ be a function that is analytic inside and on a simple closed contour C, except for a finite number of

isolated singularities (poles) inside C. If $f(z)$ has isolated singularities at  $z_1,z_2,\dots ,z_n$ inside C, then the integral of $f(z)$ around  C is given by

${\oint }_{C}f(z)dz=2\pi i\sum _{k=1}^n\text{Res}(f,z_k)$

Explanation:

${\int }_{\gamma }\frac{dx}{z(z-2)}$, where the contour $\gamma (\theta )$ is piecewise defined as:

$\gamma (\theta )=\{\begin{array}{ll}{e}^{2i\theta }& \text{for }\theta \in [0,\frac{\pi }{2}]\\ 1+2e^{2i\theta }& \text{for }\theta \in [\frac{\pi }{2},\frac{3\pi }{2}]\\ e^{2i\theta }& \text{for }\theta \in [\frac{3\pi }{2},2\pi ]\end{array}$

We will apply the Cauchy Residue Theorem to solve this integral. The integrand is

$f(z)=\frac{1}{z(z-2)}$

This function has two singularities at $z=0$  and $z=2$ . If the contour $\gamma$ encloses these singularities, the value of the integral is determined by the residues of the function at these points.

Here winding number of z = 0 is 2 and winding number of z = 2 is 1.

Residue at $z=0$ :

The residue at $z=0$ is found by multiplying $f(z)$ by $z$ and taking the limit as  $z\to 0$:

$\text{Res}(f,0)=\underset{z\to 0}{lim}z\cdot \frac{1}{z(z-2)}=\frac{1}{-2}=-\frac{1}{2}$

Residue at $z=2$:

The residue at $z=2$ is found similarly,

$\text{Res}(f,2)=\underset{z\to 2}{lim}(z-2)\cdot \frac{1}{z(z-2)}=\frac{1}{2}$

Applying the Residue Theorem:

Since the contour $\gamma$ encloses both singularities at $z=0$ and $z=2$ , the value of the integral is

I = $2\pi i\times (\text{sum of residues})$

  = $2\pi i\times (-\frac{1}{2}\times 2+\frac{1}{2}\times 1)$

  = $2\pi i(-\frac{1}{2})$ = -πi

Hence, the correct option is (3).