13. Complex Variables

Transcription

1 . Complex Variables QUESTION AND ANSWERE Complex Analysis:. The residue of the function f() = ( + ) ( ) (a) (b) (c) 6 6 at = is (d) [EC: GATE-8]. (a) d Residue at = is lim ( ) f() d d = lim d + = lim + ( ) = lim = ( + ). If f() = c c, + than + f() d is given by [EC: GATE-9] unit circle (a) π c (b) π (+ c ) (c) π jc (d) π j(+ c ). (d) ( ) + f() + c + c Let g() = = g() has a pole of order two at = d Re s(g,o) = lim ( ) g() d = lim + c ( ) = + c = g()d = π i( + c ) Page of 9

2 . The residues of a complex function X ( ) = at its poles are [EC: GATE-] ( )( ) (a), and (b), and (c), and (d), and. (c) x() has simple poles at =,,. = = = ( ) ( ) ( ) Res x, lim x lim ( )( ) Res ( x,) = lim ( ) x ( ) lim = = ( ) Res ( x,) = lim ( ) x ( ) lim = = ( ) 4. For the function of a complex variable W = ln Z (where, W = u + jv and Z = x + jy), the u = constant lines get mapped in Z-plane as [EC: GATE-6] (a) set of radial straight lines (b) set of concentric circles (c) set of confocal hyperbolas (d) set of confocal ellipses 4. Ans. (b) Given, W = loge u + iv = log e(x + iy) = log(x + y ) + i tan Since, u is constant, therefore log(x + y ) = c x + y = c Which is represented set of concentric circles. y x 5. The value of the contour integral d in positive sense is [EC: GATE-6] 4 (a) j π j π (c) j = + (b) (d) π π Page 4 of 9

3 5. (d) Let f() = = ( )( ) i i In i =, = i is a pole of order. Re s ( f,i) = lim ( i) f ( ) i = 4i π f()d = π i = 4i i= (,) (,i) (,-) -i = ME Years GATE Questions 7. i i, where i=, is given by [ME: GATE-996] π/ π (a) (b) e (c) (d) 7. (b) Page 5 of 9

4 i ilogi i = e. π Now, log i = log i + kπ+ i, k =,,,... πi =, for k =. π i i. πi i = e = e 8. The integral f ( d ) evaluated around the unit circle on the complex plane for cos f( ) is [ME: GATE-8] (a) πi (b) 4 πi (c) - π i (d) 8. (a) cos f() =, f() has a simple pole at = Re s(f,) = lim ( ) f ( ) = f()d = π i = πi 9. Assuming i = and t is a real number π / it edt is [ME: GATE-6] (a) + i (b) i (c) + i (d) i (a) π/ it π/ it e l i / π π π edt ( e ) i cos isin i = = i + = i + i = + i.. If φ(x,y) and ψ (x,y) are functions with continuous second derivatives, then φ(x,y) + i ψ (x,y) can be expressed as an analytic function of x+ i ψ (i= -), when [ME: GATE-7] Page 6 of 9

5 φ ψ φ ψ φ ψ φ ψ (a) =, = (b) =, = x x y y y x x y φ φ ψ ψ φ φ ψ ψ (c) + = + = (d) + = + = x y x y x y x y. (b) Φ (x,y) + i Ψ (x,y) is analylic, so it satisfies Cauchy-Riemann equation Φ = Ψ, Φ = Ψ x y y x. An analytic function of a complex variable = x + iy is expressed as f() = u (x, y) + i v(x, y) where i =. If u = xy, the expression for v should be [ME: GATE-9] ( x+ y) x y y x ( x y) (a) +k (b) +k (c) +k (d) +k. (c) Here u and v are analytic as f() is analytic. u,v satisfy Cauchy-Riemann equation. ux = v y (i) and uy = v x (ii) Given u = xy ux = y vy = y [by(i)] Integrating y v = + c(x) (iii) Again vx = c(x) uy = c(x) [by(ii)] x = c(x) Integreting, x c(x) = + k From (iii) we get y x v = + k + 4i. The modulus of the complex number i ( a)5 ( b) 5 ( c)/ 5 ( d )/5. (b) is. [ME: GATE-] Page 7 of 9

6 + 4i (+ 4i)(+ i) 5+ i = = = + i i ( i)(+ i) 5 + 4i = + i = 5 i CE Years GATE Questions Q6. For an analytic function, f ( x+iy ) = u( x,y ) +iv( x,y ),u is given by u= x y. The expression for v, considering K to be constant is (a) y x + K (b) 6x 6y+ K (c) 6y 6x+ K (d) 6 xy + K [CE-] Ans. (d) Exp. Cauchy Riemann equations for f ( ) = u+ iv ux = vyand uy = vx (ii) Given u= x y u = 6x u = 6x [using (i)] x y v = 6xy+ φ ( x) (iii) Differentiating w.r.t.x vx = 6y + φ '( x) uy = 6y +φ' ( x) 6y = 6y + φ' x φ '( x) = ( x) K ( ) [using (ii)] φ = (Constant) from (iii) we get v = 6xy+ K. The analytic function f() = + has singularities at [CE: GATE 9] (a) and (b) and i (c) and i (d) i and i. (d) f() = = + ( + i)( i) f()has singularities at = i, i 4. Using Cauchy s integral theorem, the value of the integral (integration being taken in counter 6 clockwise direction) d is [CE: GATE 6] c Page 8 of 9

7 (a) π π 4 4 π i π (b) 6π i (c) 6 π i (d) (a) 6 Let f() =. Here f() has a singularities at = i / i 6 i Res(f, ) = lim ( i / ). 6 i/ i = 7 i π 4 π f()dx = π i 6 = i 4π i = 4πi c 5. Consider likely applicability of Cauchy s Integral Theorem to evaluate the following integral counter clockwise around the unit circle c. [CE: GATE 5] I = sec d, c being a complex variable. The value of I will be (a) I = : singularities set = φ n + (a) I = : singularities set = { ± π ; n =,,... } (c) I = π : singularities set = { ± nπ; n =,, } (d) None of above 5. Ans. (a) sec d = d cos The poles are at = n + π π π π + π =...,,,... None of these poles lie inside the unit circle = Hence, sum of residues at poles = Singularities set = φ and I = πi [sum of residues of f() at the poles] = πi = cos (π ) 6. The value of the integral d (where C is a closed curve given by = ) is ( ) ( ) C [CE: GATE 9] π i (a) πi (b) (c) π i (d) πi 5 5 Page 9 of 9

8 6. (c) Let cos(π) f() =. ( )( ) f()has singularity at = / in C ( = ). Res(f, / ) = lim ( / )f() / cos(π) = lim. / = 5 πi f()d = 5 c 7. Which one of the following is NOT true for complex number Z and Z? Z (a) Z = ZZ Z (c) (b) Z + Z Z + Z [CE: GATE 5] Z Z Z Z (d) Z + Z + Z Z = Z + Z 7. Ans. (d) (a) is true since Z Z = Z Z Z Z = Z Z Z (b) is true by triangle inequality of complex number. (c) is not true since Z Z Z Z (d) is true since Z + Z = (Z + Z)(Z + Z) = (Z + Z)(Z + Z) = ZZ + ZZ + ZZ + ZZ (i) And Z Z = (Z + Z)(Z Z) = (Z Z ) (Z Z ) = ZZ + ZZ ZZ ZZ (ii) Adding (i) and (ii) we get Z + Z + Z Z = ZZ + ZZ = Z + Z Q4. Roots of the algebraic equation EE All GATE Questions x + x + x+ = are Page of 9

9 Ans. Exp, (a) ( +, + j, j) (b) ( +,, + ) (c) (,, ) (d) (, j, j) (d) x + x + x+ = () f = Now ( ) So, ( x+ ) is a factor of () x x x = x x+ + x+ + x+ = ( ) ( ) ( ) ( )( ) x+ x + = x =, j, + j + [EE-] Q. A point has been plotted in the complex plane, as shown in figure below. The plot of the complex number (a) y = [EE-] (b) (c) (d) Page of 9

10 Ans. (d) 8. Given X()= with > a, the residue of X() ( a ) n- at =a for n will be [EE: GATE-8] n (a) a (b) a n (c) na n (d) na n- 8. Ans. (d) X = with > a ( ) Let f ( a) ( ) ( ) n = = ( ) Re sidue of F at =a X ( a) ( ) F a n d = lim [( a) F( ) ]! d a d d n = lim ( ) d a d n = na 9. For the equation, s - 4s + s + 6 = The number of roots in the left half of s-plane will be (a) ero (b) one (c) two (d) three [EE: GATE-4] 9. Ans. (c) Constructing Routh-array S S -4 6 S S 6 Number of sign changes in the first column is two, therefore the number of roots in the left half s- plane is. The algebraic equation [EE: GATE-6] 5 4 F( s) = s s + 5s 7s + 4s + is given F( s) = has (a) a single complex root with the remaining roots being real (b) one positive real root and four complex roots, all with positive real parts Page of 9

11 (c) one negative real root, two imaginary roots, and two roots with positive real parts (d) once positive real root, two imaginary roots, and two roots with negative real parts. Ans. (c) 5 F( s ) =s -s -7s +4s= we can solve it by making Routh Hurwit array. 5 s s 7 s 8/ / s 5 s s We can replace st element of s by. If we observe the st column, sign is changing two times. So we have two poles on right half side of imaginary Axis and 5s += So, s =± j and pole on left side of imaginary axis. 6. d The value of where C is the contour i / = is C ( + ) [EE: GATE-7] (a) π i (b) π (c) tan (d) π i tan 6. Ans (b) IE All GATE Questions. Consider the circle 5 5i = in the complex plane (x, y) with = x + iy. The minimum distance from the origin to the circle is [IE: GATE-5] (a) 5 (b) 54 Page of 9

12 (c) 4 (d) 5. (a) 5 5i = (5 + 5i) = represents a circle of radius. and center (5,5) From figure, OP = = 5 Y P (5,5i) Q (,) X OQ is minimum distance from the origin. OQ = 5 as PQ = radius =.. Let =, where is a complex number not equal to ero. Then is a solution of [IE: GATE-5] (a) = (b) = (c) 4 = (d) 9 =. Ans. (c) Given, = 4 = (on multiplying both side) Now by hit and trial method we see the solution being Z 4 =. The value of the integral of the complex function [IE: GATE-6] s + 4 f(s) = (s + ) (s + ) Along the path s = is (a) πj (b) 4πj (c) 6πj (d) 8πj. (c) s + 4 Given f(s) = (s + )(s + ) f(s)has singularities at s =, which are inside the given circle Page 4 of 9

13 s = Re s(f, ) = lim (s + )f(s) =. s s Re s(f, ) = lim (s + )f(s) =. s= f(s)ds = π j ( + ) = 6πj 4. Let j =. Then one value of j j is [IE: GATE-7] (a) j (b) (c) π π (d) e 4. (d) same as Q.7 5. The polynomial p(x) = x 5 + x + has [IE: GATE-7] (a) all real roots (b) real and complex roots (c) real and 4 complex roots (d) all complex roots 5. (c) 5 Given f(x) = x + x +. P( + x) =+++ (Taking only sign of each term) P(x)has no +ve real roots. P(-x)= (Taking only sign of each term) P(x)has one ve real root As, P(x) of degree 5.So other four roots are complex. sin 6. For the function of a complex variable, the point = is [IE: GATE-7] (a) a pole of order (b) a pole of order (c) a pole of order (d) not a singularity 6. (b) sin As.lim =. Therefore the function has = is a pole of order. 7. It is known that two roots of the nonlinear equation x 6x + x 6 = are and. The third root will be [IE: GATE-8] (a) j (b) j (c) (d) 4 7. (c) Page 5 of 9

14 Let third root be α. of x 6x + x 6 = Then + +α = 6 α = 8. If = x + jy, where x and y are real, the value of e j is [IE: GATE-9] (a) (b) e x + y (c) e y (d) e y 8. (d) e = e = e = e e = e e j j(x+ jy) y+ jx y jx y jx = > R = y y ix e e,for all y and e 9. One of the roots of the equation x = j, where j is the positive square root of, is [IE: GATE-9] (a) j (b) + j (c) j (d) j 9. (b) x = j jπ x = e jπ π π 6 x = e = cos + jsin j.. The root mean squared value of x(t) = + sin (t) cos (t) is [IE: GATE-8] (a) (b) 8 (c) (d). Ans. (d) x(t) = + sin t cos t x(t) = + sin t sin t Root mean square value = + + =. Contour C in the adjoining figure is described by x + y = The value of d (Note : j = ) [IE: GATE-].5.5j C Page 6 of 9

15 y plane x C (a) π j (b) π j (c) 4π j (d) 4π j. (d) + 8 ( + 8) Let f() = =.5.5j j f() has a singularity at =j which is inside the given circle x + y = 6. Res(f,j) = lim ( j)f() = j f()ds = π j ( ) = 4πj Page 7 of 9