# 4-Introduction to Complex variables-13-Jul-2020Material_II_13-Jul-2020_Comp (1).pdf

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Complex Analysis Problems with solutions
Contents Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.1 Basic algebraic and geometric properties 7 1.2 Modulus 10 1.3 Exponential and Polar Form, Complex roots 13 2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.1 Basic notions 19 2.2 Limits, Continuity and Differentiation 27 2.3 Analytic functions 31 2.3.1 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3 Complex Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.1 Contour integrals 39 3.2 Cauchy Integral Theorem and Cauchy Integral Formula 43 3.3 Improper integrals 56 4 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4.1 Taylor and Laurent series 59 4.2 Classification of singularities 68
4.3 Applications of residues 74 4.3.1 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
1. Complex Numbers 1.1 Basic algebraic and geometric properties 1. Verify that (a) 2 - i - i 1 - 2 i = - 2 i (b) ( 2 - 3 i )( - 2 + i ) = - 1 + 8 i Solution. We have 2 - i - i 1 - 2 i = 2 - i - i + 2 = - 2 i , and ( 2 - 3 i )( - 2 + i ) = - 4 + 2 i + 6 i - 3 i 2 = - 4 + 3 + 8 i = - 1 + 8 i . 2. Reduce the quantity 5 i ( 1 - i )( 2 - i )( 3 - i ) to a real number. Solution. We have 5 i ( 1 - i )( 2 - i )( 3 - i ) = 5 i ( 1 - i )( 5 - 5 i ) = i ( 1 - i ) 2 = i - 2 i = 1 2
8 Chapter 1. Complex Numbers 3. Show that (a) Re ( iz ) = - Im ( z ) ; (b) Im ( iz ) = Re ( z ) . Proof. Let z = x + yi with x = Re ( z ) and y = Im ( z ) . Then Re ( iz ) = Re ( - y + xi ) = - y = - Im ( z ) and Im ( iz ) = Im ( - y + xi ) = x = Re ( z ) . 4. Verify the associative law for multiplication of complex numbers. That is, show that ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 ) for all z 1 , z 2 , z 3 C . Proof. Let z k = x k + iy k for k = 1 , 2 , 3. Then ( z 1 z 2 ) z 3 = (( x 1 + y 1 i )( x 2 + y 2 i ))( x 3 + y 3 i ) = (( x 1 x 2 - y 1 y 2 )+ i ( x 2 y 1 + x 1 y 2 ))( x 3 + y 3 i ) = ( x 1 x 2 x 3 - x 3 y 1 y 2 - x 2 y 1 y 3 - x 1 y 2 y 3 ) + i ( x 2 x 3 y 1 + x 1 x 3 y 2 + x 1 x 2 y 3 - y 1 y 2 y 3 ) and z 1 ( z 2 z 3 ) = ( x 1 + y 1 i )(( x 2 + y 2 i ))( x 3 + y 3 i )) = ( x 1 + y 1 i )(( x 2 x 3 - y 2 y 3 )+ i ( x 2 y 3 + x 3 y 2 )) = ( x 1 x 2 x 3 - x 3 y 1 y 2 - x 2 y 1 y 3 - x 1 y 2 y 3 ) + i ( x 2 x 3 y 1 + x 1 x 3 y 2 + x 1 x 2 y 3 - y 1 y 2 y 3 ) Therefore, ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 ) 5. Compute (a) 2 + i 2 - i ; (b) ( 1 - 2 i ) 4 . Answer: (a) ( 3 + 4 i ) / 5, (b) - 7 + 24 i .
1.1 Basic algebraic and geometric properties 9 6. Let f be the map sending each complex number z = x + yi -→ x y - y x Show that f ( z 1 z 2 ) = f ( z 1 ) f ( z 2 ) for all z 1 , z 2 C .
7. Use binomial theorem ( a + b ) n = n 0 a n + n 1 a n - 1 b + ... + n n - 1 ab n - 1 + n n b n = n k = 0 n k a n - k b k to expand (a) ( 1 + 3 i ) 2011 ; (b) ( 1 + 3 i ) - 2011 . =
10 Chapter 1. Complex Numbers Similarly, ( 1 + 3 i ) - 2011 = 1 1 + 3 i 2011 = 1 - 3 i 4 ! 2011 = 1 4 2011 2011 k = 0 2011 k ( - 3 i ) k = 1 4 2011 1005 m = 0 2011 2 m ( - 3 ) m - i 4 2011 1005 m = 0 2011 2 m + 1 ( - 3 ) m 3 .