Complex Variables MCQ Quiz - Objective Question with Answer for Complex Variables - Download Free PDF

Explanation:

Let γ be the positively oriented circle in the complex plane given by {z ∈ ℂ: |z – 1| = 1}.

Then ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{z^3-1}}$ 

Concept - 

Cauchy Integral Formula - 

Let f(z) be analytic in a region D and let C be a closed curve in D. If a is any point in D, then 

${\displaystyle {\int }_C\frac{f(z)}{z-a}dz=2\pi i.f(a).\eta (\gamma :a)}$

here $\eta (\gamma :a)$ is a winding number. The winding number measures the number of times a path (counter-clockwise) winds around a point.

Explanation -

We have γ = {z ∈ ℂ: |z – 1| = 1}.

Now ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{z^3-1}}$

= ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{(z-1)(z^2+z+1)}}$

= ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{1/(z^2+z+1)}{(z-1)}dz}$

Now using Cauchy Integral formula we get -

= ${\displaystyle \frac{1}{2\pi i}\times 2\pi i\times f(1)={\displaystyle \frac{1}{2\pi i}\times 2\pi i\times \frac{1}{3}=\frac{1}{3}}}$

Hence option(ii) is correct.

Concept:

The limit point of the set of singular points of a complex function is a non-isolated essential singularity of the function.

Explanation:

f(z) = cot$\frac{1}{z}$ = $\frac{\cos \frac{1}{z}}{\sin \frac{1}{z}}$

f(z) has singularity at

sin$\frac{1}{z}$ = 0 i.e., $\frac{1}{z}$ = nπ i.e., z = $\frac{1}{n\pi }$, n ∈ $\mathbb{Z}$

Now, $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n\pi }$ = 0

Since limit point of the set of singular points is 0 so f(z) has essential singularity at z = 0

(4) is correct

Concept -

If ${\cos }^{-1}(cos\theta +isin\theta )=x+iy$ then 

real part of ${\cos }^{-1}(cos\theta +isin\theta )=x=si{n}^{-1}(sin\theta )$

imaginary part of ${\cos }^{-1}(cos\theta +isin\theta )=y=log[\sqrt{1+sin\theta }-\sqrt{sin\theta }]$

Explanation -

We have ${\cos }^{-1}\left(\frac{3-2i}{3+2i}\right)=x+iy$  

Now solve $\left(\frac{3-2i}{3+2i}\right)$

Now use the conjugate of 3 - 2i

So $\left(\frac{3-2i}{3+2i}\right)=\left(\frac{3-2i}{3+2i}\right)\times \left(\frac{3-2i}{3-2i}\right)$

= $\left(\frac{(3-2i{)}^2}{(3+2i)(3-2i)}\right)=\frac{9-4-12i}{9+4}=\frac{5-12i}{13}=\frac{5}{13}-\frac{12i}{13}$

So we get ${\cos }^{-1}(\frac{5}{13}-\frac{12i}{13})=x+iy$

Use now the formula -

imaginary part of ${\cos }^{-1}\left(\frac{3-2i}{3+2i}\right)$= y = $log[\sqrt{(1-\frac{12}{13})}-\sqrt{-\frac{12}{13}}]$

⇒ $y=log[\sqrt{(\frac{1}{13})}-i\sqrt{\frac{12}{13}}]$

Explanation:

Let, f(z) = u + iv

u = cosh z cos y

= $\left(\frac{e^x+e^{-x}}{2}\right)\cos y$

$u=\left(\frac{e^x+e^{-x}}{2}\right)\cos y$

by Mine Thomson method

$u(x,y)=\left(\frac{e^x+e^{-x}}{2}\right)\cos y$

Partial differential w.r.t. x, y

$u_x=\left(\frac{e^x-e^{-x}}{2}\right)\cos y;u_y=\left(\frac{e^x+e^{-x}}{2}\right)(-\sin y)$

$u_x=\left(\frac{e^x-e^{-x}}{2}\right)\cos y{u}_y=-\frac{(e^x+e^{-x})}{2}\sin y$

$u_x(z,0)=\frac{e^z-e^{-z}}{2}=\sinh z{u}_y(z,0)=-0$

f(z) = ∫(u_x - iu_y)dz + c

= ∫(sinhz - a)dz + c

= ∫sinh z + c

= cos h z + c

Explanation:

Consider the function $f(z)=\frac{\sin (z)}{z^3}$

The function sin(z) has the following Maclaurin series expansion around z = 0 :

$\sin (z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$

Now, divide this by z³ to get the Laurent series for f(z) :

$f(z)=\frac{\sin (z)}{z^3}=\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots }{z^3}$

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

Here, the Laurent series contains a term of 1/z² (a negative power of z ), but there are no negative powers beyond this.

This indicates that z = 0 is a pole of order 2.

1) A removable singularity occurs if the Laurent series has no negative powers of z .

Since f(z) has a negative power 1/z² , this statement is incorrect.

2) A simple pole occurs if the Laurent series has exactly one term of the form 1/z.

Since the Laurent series of f(z) contains a term of 1/z², this statement is incorrect.

3) A pole of order 2 occurs if the Laurent series contains a term of 1/z² and no terms with higher negative powers.

This is exactly the case for f(z), so this statement is correct.

4) An essential singularity occurs if the Laurent series contains infinitely many negative powers of z.

Since the Laurent series of f(z) has only a single term with a negative power 1/z², this statement is incorrect.

Thus, the correct answer is (3)

Concept:

Pole:

The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.

When the order of a pole is 1, it is known as a simple pole.

Residue:

If f(z) has a simple pole at z = a, then

$Resf(a)=\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$

If f(z) has a pole of order n at z = a, then

$Res\left(atz=a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Calculate:

Given:

$f\left(z\right)=\frac{z^2}{z^2+a^2}$

For calculating pole:

z² + a² = 0

∴ (z + ia)(z - ai) = 0

∴ z = ai, -ai.

∴ z has simple pole at z = ai and -ai.

Residue:

If f(z) has a simple pole at z = a, then

$Resf(a)=\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$

For pole at z = ai

$Resf(ai)=\underset{z\to ai}{lim}\left(z-ai\right)\left(\frac{z^2}{z^2+a^2}\right)$

$Resf(ai)=\underset{z\to ai}{lim}\left(z-ai\right)\left(\frac{z^2}{(z-ai)(z+ai)}\right)$

$Resf(ai)=\frac{(ai{)}^2}{2ai}\Rightarrow \frac{ai}{2}$

For pole at z = -ai

$Resf(-ai)=\underset{z\to -ai}{lim}\left(z+ai\right)\left(\frac{z^2}{(z-ai)(z+ai)}\right)$

$Resf(-ai)=\frac{(-ai{)}^2}{-2ai}\Rightarrow -\frac{ai}{2}$

∴ z has a simple pole at z = ai and $\frac{ia}{2}$ is a residue at z = ia of f(z)

Concept:

Laurentz Series is obtained by the arrangement and manipulation of standard series or expansions, i.e.

(1 - x)^-1 = 1 + x + x² + x³ + …… |x| < 1

(1 + x)^-1 = 1 – x + x² – x³ + ….. |x| < 1

(1 - x)^-2 = 1 + 2x + 3x² + ….. |x| < 1

(1 + x)^-2 1 – 2x + 3x² – 4x² + …. |x| < 1

Observe that in all the expansions; |x| should be less than 1.

 ∴ We need to manipulate the variable to satisfy the above condition.

Application:

Given region |z| > 2

⇒ $\frac{2}{\left|z\right|}<1$ 

$\frac{1}{\left|z\right|}<\frac{1}{2}$

This can be interpreted as $\frac{1}{\left|z\right|}<1$ 

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

$f\left(z\right)=\frac{1}{z}[\frac{1}{1-\frac{1}{z}}-\frac{1}{1-\frac{2}{z}}]$

$Since\frac{1}{1-\frac{1}{z}}={(1-\frac{1}{z})}^{-1}$

$\left|\frac{1}{z}\right|<1\Rightarrow {(1-\frac{1}{z})}^{-1}=1+\frac{1}{z}+\frac{1}{z^2}+\dots$

Similarly, we can write:

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

$\left|\frac{2}{z}\right|<1\Rightarrow {(1-\frac{2}{z})}^{-1}=1+\frac{2}{z}+{\left(\frac{2}{z}\right)}^2+\dots$

$\therefore f\left(z\right)=\frac{1}{z}[[1+\frac{1}{z}+\frac{1}{z^2}+\dots ]-[1+\frac{2}{z}+{\left(\frac{2}{z}\right)}^2+\dots ]]$

$=[[\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\dots ]-[\frac{1}{z}+\frac{2}{z^2}+\frac{4}{z^2}+\dots ]]$

Concept:

The argument of z is the angle between the positive real axis and the line joining the point to the origin.

if a complex function is given by, z = x +iy, then the argument of z = arg(z) = ${\tan }^{-1}\left(\frac{y}{x}\right)$

 Calculations: 

Given, the complex number $\sqrt{-1}$

⇒ $z=\sqrt{-1}=i$ = x + iy

By comparing,

⇒ x = 0 and y = 1

⇒ z lies in the first quadrant.

Hence, arg(z) = $ta{n}^{-1}(\frac{y}{x})$

⇒arg(z) = $ta{n}^{-1}(\frac{1}{0})=\frac{\pi }{2}$

⇒ arg(z) = $\frac{\pi }{2}$ 

Hence, the argument of the complex number  $z=\sqrt{-1}=i$ is $\frac{\pi }{2}$

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Application:

${\int }_C\frac{z^2+1}{z^2-2z}dz$

$={\int }_C\frac{z^2+1}{z\left(z-2\right)}dz$

The simple poles are: z = 0, 2

The given region is a unit circle.

The residue at z = 2 is zero as it lies outside the given region.

The reside at z = 0, is given by

$=\underset{z\to 0}{lim}z\frac{z^2+1}{z\left(z-2\right)}dz=-\frac{1}{2}$

Cauchy-Riemann equations:

Rectangular form:

f(z) = u(x, y) + f v(x, y)

f(z) to be analytic it needs to satisfy Cauchy Riemann equations

ux = vy, uy = -vx

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

Polar form:

f(z) = u(r, θ) + f v(r, θ)

$u_r=\frac{1}{r}{v}_{\theta }$ and uθ = -rvr

​​${\displaystyle \frac{\partial u}{\partial r}}={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial v}{\partial \theta }}\text{and}{\displaystyle \frac{\partial v}{\partial r}}=-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u}{\partial \theta }}$

$\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{f}{\left(\mathrm{z}\right)}_{\mathrm{z}=\mathrm{a}}=\frac{1}{\left(\mathrm{n}-1\right)!}\frac{{\mathrm{d}}^{\mathrm{n}-1}}{\mathrm{d}{\mathrm{z}}^{\mathrm{n}-1}}{\left(\mathrm{z}-\mathrm{a}\right)}^{\mathrm{n}}{\mathrm{f}\left(\mathrm{z}\right)]}_{\mathrm{z}=\mathrm{a}}$

Here we have n = 2 and a = 2

Thus Res $\mathrm{f}{\left(\mathrm{z}\right)}_{\mathrm{z}=2}=\frac{1}{\left(2-1\right)!}\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}{\left[{\left(\mathrm{z}-2\right)}^2\frac{1}{{\left(\mathrm{z}-2\right)}^2{\left(\mathrm{z}+2\right)}^2}\right]}_{\mathrm{z}=2}$

$\begin{array}{l}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}{\left[\frac{1}{{\left(\mathrm{z}+2\right)}^2}\right]}_{\mathrm{z}=2}={\left[-\frac{2}{{\left(\mathrm{z}+2\right)}^3}\right]}_{\mathrm{z}=2}\\ =-\frac{2}{64}=-\frac{1}{32}\end{array}$

Concept:

For a complex number z = x + iy, 

the modulus is given by

$|z|=\sqrt{x^2+y^2}$

the argument is given by 

$\theta =\arg z={\tan }^{-1}\left(\frac{y}{x}\right)$

Given:

$z=\frac{1+i}{1-i}=\frac{\left(1+i\right)\left(1+i\right)}{\left(1+i\right)\left(1-i\right)}=\frac{{\left(1+i\right)}^2}{1-i^2}$

$\Rightarrow z=\frac{1+2i+i^2}{1+1}=\frac{2i}{2}=i$ = 0 + 1i

∴ x = 0, y = 1

And Arg (z) = Arg (i) = θ 

Concept:

The cube roots of unity are 1, ω and ω2 

where,

 $\omega =\frac{-1+i\sqrt{3}}{2}and{\omega }^2=\frac{-1-i\sqrt{3}}{2}$

$1+{\omega }^n+{\omega }^{2n}=\{\begin{array}{c}0,ifnisnotmultipleof3\\ 3,ifnismultipleof3\end{array}$

ω3 = 1

1 + ω + ω2 = 0

ω3n = 1

​Calculation:

Given:

Δ = $\left[\begin{array}{ccc}1& \omega & {\omega }^{2n}\\ {\omega }^2& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right]$

on solving determinant we get,

Δ = 1 (ω²ⁿ ωⁿ - 1) - ω (ω2 ωn - ω2n) + ω2n (ω2 - ω2n ω²n)

Δ = (ω³n - 1) - (ω3 ωn - ω2n ω) + (ω2 ω2n - ω⁶n )

Since ω3 = 1, ω3n = 1, ω6n = 1

Δ = 0 - ωn + ω2n ω + ω2 ω2n - 1

Δ = -1 - ωn +ω2n (ω + ω2)

Since 1 + ω + ω2 = 0 ⇒ ω + ω2 = - 1

Δ = -1 - ωn - ω2n 

If n is not a multiple of 3 then:

Δ = -1 - ωn - ω2n  = 0

Concept:

If two functions u and v satisfy Cauchy-Riemann equations, then they are said to be harmonic conjugates with respect to each other.

Cauchy-Riemann equations are 

vy = ux

vx = - uy

Calculation:

Given u = x² – y², let v be the harmonic conjugate.

By Cauchy-Riemann equations,

v_y = u_x = 2x; v_x = - u_y = - (-2y) = 2y;

We have dv = v_x dx + v_y dy 

⇒ dv = 2y dx + 2x dy = d(2xy)

⇒ v = 2xy + k or v = 2xy

∴ The conjugate harmonic function is 2xy

Concept:

Taylor series:

$f(z)=f(a)+\frac{f^{\prime }(a)}{1!}(z-a)+\frac{f^{″}(a)}{2!}(z-a{)}^2+...$

Sum of GP: $f\left(z\right)=\frac{a}{1-r}$

Calculation:

f(z) = 1 + (1 – z) + (1 – z)² + ...∞ 

The above series is in the form of G.P.

a = 1, r = (1 – z)

$f\left(z\right)=\frac{a}{1-r}=\frac{1}{1-\left(1-z\right)}=\frac{1}{z}$