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

Concept:

Taylor series expansion of f(x) about x = 0 is

f(x) = \(f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\frac{x^4}{4!}f^{(4)}(0)+...\)

Explanation:

f(x) = ln(cosh x) ⇒ f(0) = ln(cosh 0) = 0

f'(x) = tanh x ⇒ f'(0) = 0

f''(x) = sech²x ⇒ f''(0) = 1

f'''(x) = 2sech x (- sech x tanh x) = - 2sech²x tahn x ⇒ f'''(0) = 0

f⁽⁴⁾(x) = - 4sech x(- sech x tanh x)tanh x - 2sech²x sech2x ⇒ f(4)(0) = - 2 

Hence using Taylor series expansion at x = 0 we get

f(x) = \(0+x\times0+\frac{x^2}{2!}\times1+\frac{x^3}{3!}\times0+\frac{x^4}{4!}\times(-2)+...\) = \(\frac12x^2-\frac1{12}x^4+...\)

(2) is correct

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

\(\dfrac{dw}{dz}=-i\dfrac{\partial w}{\partial y}\)

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

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.

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 \(\sum a_nx^n\) then the series converges when |x| < R

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

So interval of convergence is (- R, R)

(1) is correct

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 -

we have \(\rm f(z)=|\operatorname{Re}(z) \operatorname{Im}(z)|^{\frac{1}{2}}\)

Let assume \(z= x+iy\) here Re(z) = x and Im(z) = y

so we get - \(f(z) = |xy|^{\frac{1}{2}}\)

Clearly \(u(x,y) =|xy|^{\frac{1}{2}} \ \ and \ \ v(x,y) =0\)

we know that the C-R equations - \(u_x=v_y \ \ and \ \ u_y = -v_x\)

So f(z) satisfy C-R equations at z = 0. but it is not differentiable at 0.

Hence option (iii) is correct.

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

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