analytic functions mcq +notes
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8/17/2019 Analytic Functions MCQ +Notes
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SRM UNIVERSITYRAMAPURAM PART- VADAPALANI CAMPUS, CHENNAI – 600 026
Department of MathematicsSub Title: ADVANCED CALCULUS AND COMPLEX ANALYSIS
Sub Code:15 MA102
Unit -IV - ANALYTIC FUNCTIONS
Part – A 1. Cauchy-Riemann equations are
(a) y x vu and x y vu (b) y x vu and x y vu (c) x x vu and y y vu
(d) y x vu and x y vu Ans : (a)
2. If ivu z f )( in polar form is analytic then r u
is
(a)
v
(b)
vr (c)
v
r
1 (d)
v
Ans : (c)
3. If ivu z f )( in polar form is analytic then
u
is
(a)r
v
(b)r
v
r
1 (c)
r
v (d)
r
vr Ans : (d)
4. A function u is said to be harmonic if and only if
(a) 0 yy xx uu (b) 0 yx xy uu (c) 0 y x uu (d) 022
y x uu Ans : (a)
5. A function )( z f is analytic function if
(a) Real part of )( z f is analytic (b) Imaginary part of )( z f is analytic
(c) Both real and imaginary part of )( z f is analytic (d) none of the above Ans : (c)
6. If u and v are harmonic functions then ivu z f )( is(a) Analytic function (b) need not be analytic function(c) Analytic function only at 0 z (d) none of the above Ans : (a)
7. If )()( cybxiay x z f is analytic then a,b,c equals to
(a) 1c and ba (b) 1a and bc (c) 1b and ca (d) 1 cba Ans : (a)
8. A point at which a function ceases to be analytic is called a(a) Singular point (b) Non-Singular point (c) Regular point (d) Non-regular point
Ans : (a) 9. The function ||)( z z f is a non-constant
(a) analytic function (b) nowhere analytic function (c) non-analytic function (d) entire functionAns : (b)
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10. A function v is called a conjugate harmonic function for a harmonic function u in whenever(a) ivu f is analytic (b) u is analytic (c) v is analytic (d) ivu f is analytic
Ans : (a)
11. The function 3223)( cybxy yax xiy x f is analytic only if
(a) 3,3 bia and ic (b) 3,3 bia and ic (c) 3,3 bia and ic
(d) 3,3 bia and ic Ans : (c)
12. There exist no analytic functions f such that(a) Re x y z f 2)( (b) Re x y z f 2)( 2 (c) Re 22)( x y z f (d) Re x y z f )(
Ans : (b)
13. If ye ax cos is harmonic, then a is(a) i (b) 0 (c) -1 (d) 2 Ans : (a)
14. The harmonic conjugate of 23 32 xy x x is
(a)323 y y x x (b)
3232 y y x y (c)323 y y x y (d) 3232 y y x y Ans : (b)
15. The harmonic conjugate of )1(2),( y x y xu is
(a) C x y x 222
(b) C y y x 222
(c) C y y x 222
(d) C y y x 222
Ans : (d)
16. harmonic conjugate of xe y xu y cos),( is
(a) C ye x cos (b) C ye x sin (c) C xe y sin (d) C xe y sin Ans : (d)
17. If the real part of an analytic function )( z f is ,22 y y x then the imaginary part is
(a) xy2 (b) xy x 22
(c) y xy2 (d) x xy2 Ans : (d)
18. If the imaginary part of an analytic function )( z f is ,2 y xy then the real part is
(a) y y x 22
(b) x y x 22 (c) x y x 22 (d) y y x 22 Ans : (c) 19. z z f )( is differentiable
(a) nowhere (b) only at 0 z (c) everywhere (d) only at 1 z Ans : (a)
20. 2
)( z z f is differentiable
(a) nowhere (b) only at 0 z (c) everywhere (d) only at 1 z Ans : (b)
21. 2
)( z z f is
(a) differentiable and analytic everywhere(b) not differentiable at 0 z but analytic at 0 z (c) differentiable at 1 z and not analytic at 1 z only(d) differentiable at 0 z but not analytic at 0 z Ans : (d)
22. If,0,0
;0,)()( 22
zif
zif y x xy
z f then )( z f is
(a) continuous but not differentiable at 0 z (b) differentiable at 0 z (c) analytic everywhere except at 0 z (d) not differentiable at 0 z Ans : (d)
23. ze z f )( is analytic
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(a) only at 0 z (b) only at i z (c) nowhere (d) everywhere Ans : (d)
24. )sin(cos yi ye x is
(a) analytic (b) not analytic (c) analytic when 0 z (d) analytic when i z Ans : (b)
25. If )( z f is analytic, then )( z f is
(a) analytic (b) not analytic (c) analytic when 0 z (d) analytic when 1 z Ans : (a)
26. The points at which)23(
)()( 2
2
z z
z z z f is not analytic are
(a) 0 and 1 (b) 1 and -1 (c) i and 2 (d) 1 and 2 Ans : (d)
27. The points at which1
1)( 2 z
z f is not analytic are
(a) 1 and -1 (b) i and -i (c) 1 and i (d) -1 and -i Ans : (b)
28. The harmonic conjugate of 22log y xu is
(a) 22 y x
x (b) 22
y x
y (c)
y
x1tan (d)
x
y1tan Ans : (d)
29. If ),2()( z z z f then )1( i f (a) 0 (b) i (c) -i (d) 2 Ans : (b)
30. If z z f )( then )43( i f
(a) 0 (b) 5 (c) -5 (d) 12 Ans : (b)
31. Critical points of the bilinear transformationdzc
bzaw are
(a) a,c (b) ,d
c (c) ,
d
c (d) None of these Ans : (c)
32. The points coincide with their transformations are known as(a) fixed points (b) critical points (c) singular points (d) None of these Ans : (a)
33. dzcbzaw is a bilinear transformation when
(a) 0bcad (b) 0bcad (c) 0cd ab (d) None of these Ans : (b)
34. z
w1
is known as
(a) inversion (b) translation (c) rotation (d) None of these Ans : (a) 35. zw is known as
(a) inversion (b) translation (c) rotation (d) None of these Ans : (b) 36. A translation of the type zw where and are complex constants, is known as a
(a) translation (b) magnification (c) linear transformation (d) bilinear transformationAns : (c)
37. A mapping that preserves angles between oriented curves both in magnitude and in sense is called a/an .....mapping.(a) informal (b) isogonal (c) conformal (d) formal Ans : (c)
38. The mapping defined by an analytic function )( z f is conformal at all points z except at points where
(a) 0)(' z f (b) 0)(' z f (c) 0)(' z f (d) 0)(' z f Ans : (a)
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39. The fixed points of the transformation 2 zw are(a) 0,1 (b) 0,-1 (c) -1,1 (d) – i,i Ans : (a)
40. The invariant points of the mapping z
zw
2 are
(a) 1,-1 (b) 0,-1 (c) 0,1 (d) -1,-1 Ans : (c)
41. The fixed points of11
z z
w are
(a) 1 (b) i (c) 0,-1 (d) 0,1 Ans : (b)
42. The mapping z
zw 1
transforms circles of constant radius into
(a) confocal ellipses (b) hyperbolas (c) circles (d) parabolas Ans : (a)
43. Under the transformations ,1
zw the image of the line
4
1 y in z-plane is
(a) circle 0422
vvu (b) circle 422
vu (c) circle 222
vu (d) none of theseAns : (a)
44. The bilinear transformation that maps the points ,,0 i respectively into ,1,0 is w
(a) z
1 (b) – z (c) – iz (d) iz Ans : (c)
45. The bilinear transformation which maps the points 1,0,1 z z z of z - plane into 1,0, wwiw ofw plane respectively is(a) izw (b) zw (c) )1( ziw (d) none of these Ans : (a)
Part – B 1. Show that the function f (z) = is no where differentiable.
Solution: Given u+iv = x-iyu=x v=-yux =1 vx =-1uy =0 vy =-1
ux vy
C-R equations are not satisfied.
f (z) = is no where differentiable.
2. Show that f (z) = is differentiable at z=0 but not analytic at z=0.
Solution: Let= z =
v=0
ux =2x vx =0uy =2y v y = 0ux = vy and u y = - vx are not satisfied everywhere except at z=0
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So f (z) may be differentiable only at z=0. Now u x,vx,u y,vy are continuous everywhere and inparticular at (0,0).
3. Test the analyticity of the function w=sin z.Solution: w=f (z) =sin zu+iv = sin(x+iy)=sin x cosiy+ cos x siniy
= sin x coshy+i cos x sinhyu= sin x cushy v= cos x sinhy
ux = cosx cushy v x = -sinx sinhyuy = sinx sinhy v y = cosx cushyux = vy and u y = - vx
C-R equations are satisfied.
The function is analytic.
4. Verify the function 2xy+i( ) is analytic or not .
Solution: u=2xy v=ux = 2y v x = 2xuy = 2x vy = -2y
ux vy and u y - vx
C-R equations are not satisfied.
The function is not analytic.
5. Test the analyticity of the function f (z) = .
Solution: f (z) =
u+iv = = = (cosy+isiny)
u= cosy v= siny
ux = cosy v x = siny
uy = siny v y = cosy
ux = vy and u y = - vxThe function is analytic.
6. If u+iv = is analytic, show that v-iu and –v+iu are also analytic.Solution: Given u+iv is analytic.
C-R equations are satisfied.
i.e. u x = vy ------------------- (1) and u y = - vx------------------------------(2)To prove v-iu and –v+iu are also analyticFor this, we have to show that
(i) ux = vy and -u y = vx (ii) ux = vy and u y = - vx
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These results follow directly from (1) & (2) by replacing u by v and –v and v –u and u respectively.v-iu and –v+iu are analytic.
7.Give an example such that u and v are harmonic but u+iv is not analytic.Solution: Consider the function w= = x-iy
u=x v=-y
ux vy , The function f(z) is not analytic. But and gives u and v are
harmonic. 8.If f (z) = u(x,y) +v(x,y) is an analytic function. Then the curves u(x,y) = c 1and v(x,y) =c 2 where c 1andc2 are constants are orthogonal to each other.
Solution: If u(x,y) = c 1 , then du = 0But by total differential operator we have
du =
(Say)
Similarly, for the curve v(x,y) =c 2 we have
(Say)
For any curve gives the slope, Now the product of the slopes is
u(x,y) = c 1and v(x,y) =c 2 intersect at right angles (i.e) they are orthogonal to
each other.9.Find the analytic region of f (z) =
Solution: Given f (z) =
u= v=
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Now u x = vy and u y = - vx2 =2 - 2 = -2
x-y=1 x-y=1
Analytic region of f (z) is x-y=1
10.Find a function w such that w=u+iv is analytic, if u= .
Solution: Given u=
= 0-i
f (z) = -i
11. Prove that u= satisfies Laplace’s equation. Solution: Given u=
u satisfies Laplace’s equation.
12. If u=log ( ) find v and f (z) such that f (z) = u+iv is analytic.
Solution: Given u=log ( )
=
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f (z) = 2log z +cTo find the conjugate harmonic v
We know that dv =
= - [by C – R equations]
dv = dx
Integrating
V = 2 +c
13. Find the critical points for the transformation
Solution: Given
2w
w
Critical points occur at
Also
The critical points occur at
=0
z = and z =
The critical points occur at z = , and .
14. Find the image of the circle under the transformation w=3z.
Solution: w=3zu+iv = 3(x+iy)u=3x v=3y
x= y=
Given
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.
maps to a circle in w- plane with centre at the origin and radius 6.
15. Find the fixed points for the following transformation w
Solution: Fixed points are obtained fromf (z) = z
z =
Z = are the fixed points.
Part – C
1. If f(z) is an analytic function of z, prove that
(i) =0
(ii)
(iii)
Proof: If z = x+iy then
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=
=
(i). =
= 2
= 2
= 2
=2 = 0(ii)
=
=
=
=
=2f’ (z)
(iii). =
=
=
= 4 =
2.
Prove that the function u = satisfies laplace’s equation and find thecorresponding analytic function f (z) = u+iv.Solution: Given u =
+
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+
=
u satisfies Laplace equation.
To find f (z): u is givenStep 1:
+
Step 2:
Step3:
Integrating f (z) =
=3. Prove that the function v = is harmonic and determine the corresponding
analytic function of f(z)Solution: Given v =
Step 1:
+y
Step 2:
Step3:
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Integrating f (z) = -z
To prove v is harmonic
+y
=
4. Prove that the function u = +1 satisfies laplace’s equation and find thecorresponding analytic function f (z) = u+iv. Solution: Given u = +1
= -6x-6
u satisfies Laplace equation.
To find f (z): u is givenStep 1:
Step 2:
Step3:
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Integrating f (z) =
5. If u= find the corresponding analytic function f(z) u+iv.
Solution: Given u=
To find f (z): u is givenStep 1:
Step 2:
=
Step3:
Integrating f (z) = tan z
6. Determine the analytic function f(z)=u+iv such that v =
Solution: f (z) =u+iv ----------------------------- (1)i f(z) = iu-v ------------------------------(2)
Adding (1) and (2)
F (z) = U+iVWhere F (z) = , U= V =
Given v =
Step 1:
Step 2:
Step3:
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Integrating F (z) =
(1+i) f (z) =
7. Find the analytic function f(z) = u+iv given that
Solution: 3f (z) = 3u+3iv ---------------------- (1)2if (2) = 2iu-2v ----------------------- (2)Adding (1) and (2)(3+2i) f (z) = (3u-2v) +i (2u+3v)F (z) = U+iVWhere F (z) = (3+2i) f (z) , U= V =
Given
i.e., V =
Step 1:
Step 2:
Step3:
Integrating F (z) = i cot z
(3+2i) f (z) = i cot z
f (z)
f (z)
8. Find the bilinear transformation that maps the points z = 1, i, -1 into the points w=i, 0, -irespectively. Hence find the image of
Solution: The bilinear transformations is given by
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w+5=3z-5
w= is the required bilinear transformation.
To get the invariant points, put w=z
z=
Solving for z,
Z =
= 1
The invariant points are z = 1
10. Find the image of under the transformation.Solution: Given w = 1/z
z = x+iy and w = u+iv
And
=2
--------------------------- (1)
Substituting x and y values in equation (1), we get
This is the straight line equation in the w-plane.
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11.Show that the transformation w = 1/z transforms circles and straight line in the z-plane intocircles or straight lines in the w-plane.
Solution: w = 1/z
z = x+iy and w = u+iv
Consider the equation, ----------------------- (1)
If a equation (1) represents a circle and if a=0, it represents a straight line, substituting the
valus of x and y in (1)
------------------------------------ (2)
If d 0, equation (2) represents a circle and if d=0, it represents a straight line. The various cases
are discussed in detail.Case (i): When a d 0
Equation (1) and (2) represents circles in the z-plane and w-plane not passing through the origin.The transformation w =1/z transforms circles not passing through the origin into circles not
passing through the origin.Case (ii): When a d=0
The equation (1) is circle through the origin in z-plane and (2) is a straight line; not passingthrough the origin in the w-plane.Circles passing through the origin in the z-planes maps into the straight lines, not passing
through the origin in the w-plane.Case (iii): When a = d 0
Equation (1) represents a straight line not passing through the origin and (2) represents a circle inthe w-plane passing through the origin. Thus lines in the z-plane not passing through the originmap into circles through the origin in the w-plane.Case (iv): When a = d= 0
Equation (1) and (2) represents straight lines passing through the origin. Thus the lines through theorigin in the z- plane map into the lines through the origin in the w- plane.
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12. If u= , v= prove that u and v are harmonic functions but u+iv is not an
analytic function.
Solution: Given u= and v=
To prove u and v are harmonic
Now
u is harmonic.
Now v=
is harmonic.
Now we show that u+iv is not analytic.
Now and
It is true from the above relation.u+iv is not an analytic function.
13. Prove that u = is harmonic and find its conjugate harmonic.
Solution: Given u =
To prove
Consider u =
Differentiating this w.r.to x and y partially, we get
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u is harmonic.To find the harmonic conjugateLet v (x,y) be the conjugate harmonic. Then w = u+iv is analytic.
By C-R equations, and =
We have
dv =
dv =
dv =
Integrating, we get
V =
14. . Find the bilinear transformation that maps the points z = -1, 0, 1 into w=0, i, 3irespectively.
Solution: The bilinear transformations is given by
2w (z-1) = (w-3i) (z+1)w [2z-2-z-1] = (z+1)(-3i)
w = is the required bilinear transformation.
15. Find the bilinear transformation that maps the points z = 0, 1, into the points
w=-1,-2-i, i respectively.Solution: The bilinear transformations is given by
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Since the above relation becomes.
2w+2=-zw+izW (z+2) = iz-2
w = is the required bilinear transformation.