Complex Variables
Exam Duration: 45 Mins Total Questions : 15
In the Taylor's series expansion of ex about x=2, the coefficient of (x-2)4 is
- (a)
\(\frac { 1 }{ 4! } \)
- (b)
\(\frac { { 2 }^{ 4 } }{ 4! } \)
- (c)
\(\frac { { e }^{ 2 } }{ 4! } \)
- (d)
\(\frac { { e }^{ 4 } }{ 4! } \)
Taylor series expression of f(x) around (a-2) is
\(f[4+(x-a)]=f(x)=f(a)+\frac { x-a }{ 1! } f^{ ' }(a)+\frac { { (x-a) }^{ 2 }f^{ '' }\left( a \right) }{ 2! } +\frac { (x-a)^{ 3 } }{ 3! } f^{ ''' }\left( a \right) +\frac { (x-a)^{ 4 } }{ 4! } f^{ '''' }\left( a \right) \)
Coefficient of \((x-a)^{ 4 }is\frac { f^{ '''' }\left( a \right) }{ 4! } \)
At a=2, coefficient \({ (x-2) }^{ 4 }\)
\(\frac { f^{ '''' }\left( 2 \right) }{ 4! } =\frac { { e }^{ 2 } }{ 4! } (f^{ '''' }\left( x \right) ={ e }^{ x })\)
if \(\phi (x,y)and\psi (x,y)\) are functions with continuous second derivative, then
\(\phi (x,y)+\psi (x,y)\)
can be ecpressed as an analytic funtion of \(x+iy=(i=\sqrt { -1) } \)when
- (a)
\(\frac { \partial \phi }{ \partial x } =-\frac { \partial \psi }{ \partial x } ;\frac { \partial \phi }{ \partial y } =\frac { \partial \psi }{ \partial y } \)
- (b)
\(\frac { \partial \phi }{ \partial y } =-\frac { \partial \psi }{ \partial x } ;\frac { \partial \phi }{ \partial x } =\frac { \partial \psi }{ \partial y } \)
- (c)
\(\frac { { \partial }^{ 2 }\phi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\phi }{ { \partial y }^{ 2 } } =\frac { { \partial }^{ 2 }\psi }{ { \partial w }^{ 2 } } +\frac { \partial ^{ 2 }\psi }{ { \partial y }^{ 2 } } =1\)
- (d)
\(\frac { \partial \phi }{ \partial x } +\frac { \partial \phi }{ \partial y } =\frac { \partial \psi }{ \partial x } +\frac { \partial \psi }{ \partial y } =0\)
Given, \(\phi (x,y) and \psi (x,y)\) are continuous second derivatives function so it can be expressed as an analytic function.
Since, Cauchy-Riemann equation statistics the analytic function.
f(z)=W=\(\mu \) + iv
Then, Cauchy-Riemann equation is
\(\frac { \partial \mu }{ \partial x } =\frac { \partial v }{ \partial y } ,\frac { \partial \mu }{ \partial y } =-\frac { \partial v }{ \partial x } \)
But given \(W=\phi (x,y)+i\psi (x,y)\)
By Cauchy-Riemann equation,
\(\frac { \partial \mu }{ \partial x } =\frac { \partial v }{ \partial y } ,\frac { \partial \mu }{ \partial y } =-\frac { \partial v }{ \partial x } \)
The \(\lim _{ x\rightarrow 0 }{ \frac { sin\left[ \frac { 2 }{ 3 } x \right] }{ x } } \) is
- (a)
\(\frac { 2 }{ 3 } \)
- (b)
1
- (c)
\(\frac { 1 }{ 4 } \)
- (d)
\(\frac { 1 }{ 2 } \)
\(\lim _{ x\rightarrow 0 }{ \frac { sin\left[ \frac { 2 }{ 3 } x \right] }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { sin\left[ \frac { 2 }{ 3 } x \right] }{ 2 } } +\frac { 2 }{ 3 } \left\{ \because \lim _{ x\rightarrow 0 }{ \frac { sinx }{ x } =1 } \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =1\times \frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
The analytic function \(f(z)=\frac { z-1 }{ { z }^{ 2 }+1 } \) has singularities at
- (a)
1 and -1
- (b)
1 and 1
- (c)
1 and -i
- (d)
i and -i
Analytic function \(f(z)=\frac { z-1 }{ { z }^{ 2 }+1 } \)
\(f(z)=\frac { z-1 }{ \left( z-i \right) \left( z+i \right) } \quad [\because { i }^{ 2 }=-1]\)
So, function is not defined at x=+ i, i.e., the singularities of function is =i and -i
For an analytic function f(x+iy)=\(\mu \) (x,y) + iv(x,y), \(\mu \) is given by \(\mu \) = 3x2-3y2 expression for v considering K to be at constant is
- (a)
3y2-3x2+K
- (b)
6x-6y+K
- (c)
6x+6y+K
- (d)
6xy+K
Given \(\mu ={ 3x }^{ 2 }-{ 3y }^{ 2 }\)
\(\frac { \partial \mu }{ \partial x } =6x,\frac { \partial \mu }{ \partial y } =-6y\)
We have \(dv=\frac { \partial v }{ \partial x } dx+\frac { \partial v }{ \partial y } dy\)
\(\quad =\left( -\frac { \partial \mu }{ \partial y } \right) dx+\left( \frac { \partial \mu }{ \partial x } \right) dy\)
\(dv=6ydx+6xdy\\ \quad v=6xy+K\)
The value of the function f(x)=\(\lim _{ x\rightarrow 0 }{ \frac { { x }^{ 3 }+{ x }^{ 2 } }{ { 2x }^{ 3 }-{ 7x }^{ 2 } } } \) is
- (a)
0
- (b)
\(-\frac { 1 }{ 7 } \)
- (c)
\(\frac { 1 }{ 7 } \)
- (d)
\(\infty \)
\(f(x)=\lim _{ x\rightarrow 0 }{ \frac { { x }^{ 3 }+{ x }^{ 2 } }{ { 3x }^{ 3 }-{ 7x }^{ 2 } } =\lim _{ x\rightarrow 0 }{ \frac { x+1 }{ 2x-7 } } } \\ \quad \quad \quad =\frac { 0+1 }{ 0-7 } =-\frac { 1 }{ 7 } \)
\(\lim _{ x\rightarrow \infty }{ \frac { x-sinx }{ x+cosx } } \) equals to
- (a)
1
- (b)
-1
- (c)
\(\infty \)
- (d)
-\(\infty \)
\(\lim _{ x\rightarrow \infty }{ \frac { x-sinx }{ x+cosx } } =\lim _{ x\rightarrow \infty }{ \frac { 1-\frac { sinx }{ x } }{ 1+\frac { cosx }{ x } } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 1-0 }{ 1+0 } =1\)
The value of the \(\oint { \frac { -3z+4 }{ { z }^{ 2 }+4z+5 } } dz\) where C is the circle |z|=1 is given by
- (a)
0
- (b)
\(\frac { 1 }{ 10 } \)
- (c)
\(\frac { 4 }{ 5 } \)
- (d)
1
\(\oint { \frac { -3z+4 }{ { z }^{ 2 }+4z+5 } } dz\) where C is the circle |z|=1
Here, poles are at z=-8+i, which are outside the unit circle.
Hence, \(\oint { _{ c }\frac { -3z+4 }{ { z }^{ 2 }+4z+5 } } dz=0\)
Which of the following functions would have only odd poweers of x in its Taylor series expansion about the point x=0?
- (a)
sin (x3)
- (b)
sin (x3)
- (c)
cos (x3)
- (d)
cos (x2)
\(sin({ x }^{ 3 })={ x }^{ 3 }-\frac { ({ x }^{ 3 }{ ) }^{ 3 } }{ 3! } +\frac { ({ x }^{ 3 }{ ) }^{ 5 } }{ 5! } -\frac { ({ x }^{ 3 }{ ) }^{ 7 } }{ 7! } +...\)
\(\lim _{ \theta \rightarrow 0 }{ \frac { sin\frac { \theta }{ 2 } }{ \theta } } \) is
- (a)
0.5
- (b)
1
- (c)
2
- (d)
Not defined
\(\lim _{ \theta \rightarrow 0 }{ \frac { sin\frac { \theta }{ 2 } }{ \theta } } =\lim _{ \theta \rightarrow 0 }{ \frac { sin\frac { \theta }{ 2 } }{ \theta } } .\frac { 1 }{ 2 } =1.\frac { 1 }{ 2 } =0.5\)
Consider likely applicability of Cauchy's integral theorem to evaluate the following integral counter clockwise around the unit circle C.
\(1=\oint { _{ c } } sec\quad z\quad dz,z\) being a complex variable. The value of I will be
- (a)
l=0,singularity set \(\oint { . } \)
- (b)
l-0, singularity set \(\left\{ \pm \frac { 2n+1 }{ 2 } \pi ,n=0,1,2,... \right\} \)
- (c)
\(l=\frac { \pi }{ 2 } ,singularities\quad set=\left\{ \pm n\pi ;n=0,1,2,... \right\} \)
- (d)
None of the above
Given, \(I=\oint { _{ C } } sec\quad z\quad dz=\oint { _{ C } } \frac { 1 }{ cos\quad z } =1\\ \Rightarrow tan=\infty =tan\frac { \pi }{ 2 } \\ z=\left( n+\frac { 1 }{ 2 } \right) \pi \quad or\quad { z }_{ 0 }=\left( n+\frac { 1 }{ 2 } \right) \pi \)
So, poles are at origin
\({ z }_{ 0 }=-\frac { 3\pi }{ 2 } ,-\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } ,\frac { 3\pi }{ 2 } ,...\)
We observe that,
None of these poles lies inside the unit circle
|z|=1
Hence, the sum of residues at origin
\(=...+\left[ (-\frac { 3\pi }{ 2 } +\frac { 3\pi }{ 2 } )+(\frac { \pi }{ 2 } +\frac { \pi }{ 2 } )+... \right] =0\)
So, singularities set =\(\oint { . } \)
I=2\(\pi i\) \((\sum { { R }_{ 0 } } )\)
=2\(\pi i\) {Sum of residues at origin}
=2\(\pi i\)(0)
The residue of the function \(f(z)=\frac { 1 }{ { (z+2) }^{ 2 }(z-2) } \) at z=2 is
- (a)
\(-\frac { 1 }{ 32 } \)
- (b)
\(-\frac { 1 }{ 16 } \)
- (c)
\(\frac { 1 }{ 16 } \)
- (d)
\(\frac { 1 }{ 32 } \)
The residue at z=a is for a pole of order n as
Res \(f(a)=\frac { 1 }{ (n-1)! } \frac { { d }^{ n-1 } }{ { dz }^{ n-1 } } [(z-a)^{ n }f(z)]_{ z=a }\)
Here, n=2 (pole of order 2) and a=2
Res \(f(2)=\frac { d }{ dz } \left[ (z-2)^{ 2 }.\frac { 1 }{ { (z+2) }^{ 2 }{ (z-2) }^{ 2 } } \right] _{ z=2 }\)
\(=\frac { d }{ dz } \left[ \frac { 1 }{ (z+2)^{ 2 } } \right] _{ z=2 }=\left[ -\frac { 2 }{ (z+2)^{ 3 } } \right] _{ z=2 }=-\frac { 1 }{ 32 } \)
The Taylor series expansion of \(\frac { sin\quad x }{ x-\pi } at\quad x=\pi \quad \)is given by
- (a)
\(1+\frac { (x-\pi { ) }^{ 2 } }{ 3! } \)
- (b)
\(-1-\frac { (x-\pi { ) }^{ 2 } }{ 3! } +...\)
- (c)
\(1-\frac { (x-\pi { ) }^{ 2 } }{ 3! } \)
- (d)
\(-1+\frac { (x-\pi { ) }^{ 2 } }{ 3! } +...\)
Let f(x)=\(\frac { sin\quad x }{ x-\pi } \\ f(x)=f(\pi +x-\pi )\\ =f(x)+\frac { (x-\pi ) }{ 1! } { f }^{ ' }(x)+\frac { (x-{ \pi ) }^{ 2 } }{ 2! } { f }^{ ,, }(\pi )+...\\ Now,\quad f(\pi )=\lim _{ x\rightarrow \pi }{ \frac { sin\quad x }{ x-\pi } } =\lim _{ x-\pi }{ \frac { cos\quad x }{ 1 } =-1 } \\ { f }^{ ' }(x)=\frac { (x-\pi )cos\quad x-sin\quad x }{ (x-{ \pi ) }^{ 2 } } \\ { f }^{ ' }(x)=\lim _{ x\rightarrow \pi }{ \frac { -(x-\pi )\quad sin\quad x }{ (x-\pi { ) }^{ 2 } } } \\ \lim _{ x\rightarrow \pi }{ \frac { -(x-\pi )\quad cos\quad x-sin\quad x }{ 2 } } \\ =-1-\frac { (x-{ \pi ) }^{ 2 } }{ 3! } +...\)
The value of the contour integral \(\underset { |z-i|=2 }{ \oint { \frac { 1 }{ { z }^{ 2 }+4 } } } dz\) in positive sense is (z-i)=2
- (a)
\(\frac { \sqrt { \pi } }{ 2 } \)
- (b)
\(-\frac { \pi }{ 2 } \)
- (c)
\(-\frac { \sqrt { \pi } }{ 2 } \)
- (d)
\(\frac { \pi }{ 2 } \)
z2+4=0
z2-(2i)2=0
z=-2i lies outside the circle |z-i|=2
Simple pole of the integrand is z=2i
By Cauchy's integral formula
\(\oint { _{ |z-i|=2 } } \frac { dz }{ { z }^{ 2 }+4 } =\oint { _{ |z-i|=2 } } \frac { 1/(z+2i) }{ z-2i } dz\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =2\pi i\left( \frac { 1 }{ z+2i } \right) _{ z=2i }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi }{ 2 } \)
The Cauchy integral therom states if f(z) is analytic in a simple connected domain D, then \(\int { _{ c } } f(z)dz=0\) on every simple closed path C and Dn. The condition of analytically in this therom is
- (a)
neccessary
- (b)
necessary and sufficient
- (c)
sufficient
- (d)
None of the above
A neccessary condition.