Suggested Homework Solutions

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1 Suggested Homework Solutions Chapter Fourteen Section #9: Real and Imaginary parts of /z: z = x + iy = x + iy x iy ( ) x iy = x #9: Real and Imaginary parts of ln z: + i ( y ) ln z = ln(re iθ ) = ln r + iθ = (/) ln( ) + i arctan(y/x) NOTE: #55 Added later Section 3 #3: Evaluate z dz along indicated paths The answers are obvious using the Cauchy Theorem and just simple calculations the long way.

2 #7: Evaluate dz 8i+z along the line y = x from to. Let z = ( + i)t for t. Then dz = ( + i)dt, z = it so ( + i)dt 8i + it = + i i dt 4 + t = + i i arctan( t ) = + i 4i π = ( i)π/8 #: (a) Evaluate +i z dz along straight line from (, ) to (, ). Let z = ( + i)t, dz = ( + i)dt for t. Then z = 5t and we have 5t ( + i)dt = 5( + i) t dt = 5( + i)/3. Part (b) is just as easy. #9: Evaluate C e3z dz z ln if C is the square with vertices ± ± i. There is a simple pole at ln which is less than so inside the curve. Using the Cauchy Integral Formula, the integral equals (πi) Res = (πi) e 3z ln = (πi) e 3 ln = (πi) 8 = 6πi #3*: Evaluate C e3z dz (z ln ) 4 where C is the above square. There is a pole of order 4 at ln which is still inside the curve. Using the Generalized Cauchy Formula with n = 3, the integral equals (πi) Res = (πi) (e 3z ) ln /3! = (πi) 7 8/6 = 7πi Section 4 #6: Find all Laurent Series about the origin for z (+z). For z write ( z + z z 3 ( ) = z + 3z 4z 3 + ) = z z z z + 3 4z + 5z Note that the residue at z = is -. For z > write [ ] z 4 ( + = z ( z + z z ) = z z 3z 4z ( + ) z ) = z 4 z z 6 4 z 7

3 #9: For each function, say what type of singularity (if any) at the given point. (a) sin z z zero). (b) cos z z z = : Removable singularity (would be regular if defined to be z = : Pole of order three (numerator equals at zero). (c) z = : The numerator has a zero of order at z = - the other two (z ) 3 zeros are placed as usual on the unit circle; the denominator has an obvious zero of order three at z =. Cancelling, there is a pole of order two at z =. (d) z = : Since z ez is entire, there is a pole of order one at z =. Section 6 #3: Laurent Series for sin z z 4 about z =. sin z z 4 = z 4 ( z z 3 /3! + z 5 /5! ) = z 3 z /3! + z/5! z 3 /7! + This series converges for all z > and the residue at z = is /6. #9: Laurent Series for z 5z+6 about z =. z 5z + 6 = (z )(z 3) = z = z { + (z ) + (z ) + } = (z ) = z (z ) z (z ) (z ) This series converges in < z < and the residue at z = is -. #6: Find the residues of z z( z) at z = ; at z =. Obviously, the residue at z = is - and the residue at z = is +. #: Find the residue of z z 4 +6 at z = ( + i). The value of z is one of the four simple poles of the function so we need only evaluate z = at any of them to get the residue. Calculating 4z 3 4z 4 = ( i)/6. (+i) 3

4 Section 7 #3: Evaluate I = π dθ. 5 4 sin θ Let z = e iθ dz = ie iθ dθ = izdθ dθ = dz/(iz). Also, sin θ = (z /z)/(i). I = dz/(iz) 5 4(z /z)/(i) = dz z 5iz The roots of z 5iz = can be found by the quadratic formula to be i and i/ with i/ being the only one inside the unit circle. I = ( ) dz (z i)(z i/) = (πi)( ) Res (z i) = πi 3i/ = 3 π #: Evaluate I = I =. (4x +) 3 (4x + ) = (x + /4) = (x + i/) 3 (x i/) 3 The triple roots are at ±i/ with only the root +i/ within our standard semicircle. Using the Generalized Cauchy Formula with n = we have I = (πi) 4 Res 3 (x + i/) = πi [ ] ( 3) ( 4) 3! 4 3 (x + i/) 5 I = 3π/3 i/ = πi! i 5 4

5 4-.6: u = x u x = x + y x ( ) = y x ( ) ; u y = xy ( ) u xx = x3 6xy ( ) 3 u yy = x3 + 6xy ( ) 3 v y = u x v = y v x = u y v = y ( f(z) = x x + y + i y x + y = x iy x + y x + iy x + iy = f(z) = ( )(x + iy) = /z 5

6 4-3.4: C cosh z dz ( ln z) 5 ; C : z = Since ln <, fifth degree pole is inside the circle. For the GCIT take n=4 and note that the inside term is backward: πi [ ( ) 5 cosh z ] (4) 4! ln πi cosh z 4! ln = πi ez + e z πi [ e ln ] ln + e πi 4 ( ) ln = πi ( + ) 4 = πi = 7iπ/96 6

7 P77, #4a C z z (z + ) (z + 4) dz Poles inside figure eight at z = ±i (the double pole at z = is outside). The answer, therefore, is πi(res i Res i ). Res i = Res i = z z (z + ) = (z + i) i z z (z + ) = (z i) i The integral equals: 4 4i (i + ) (4i) = + i 3 + 4i 3 + 4i 3 + 4i = 7 + i i (i + ) ( 4i) = i 3 4i 3 4i 3 4i = 7 i 5 ( 7 + i πi 5 7 i ) = πi i 5 5 = 4π 5 7