CAtest2022solutions

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UNIVERSITY OF NEW SOUTH WALES MATH2069 MATHEMATICS 2A COMPLEX ANALYSIS TEST Show all your work. Your solutions should be uploaded together with a copy of your student ID and a declaration that your solutions are your own original work ("I declare that this submission is my own original work" and signature). Upload a SINGLE PDF FILE. You may use the lecture notes and your own notes for the test, as well as calculators or mathematical software such as maple. You may not receive any help or use internet resources. Do not discuss the test questions with anyone until Friday afternoon. Questions 1. [3 marks] (a) Sketch the curve z = 3 i + 2 e , 0 θ π . (b) Sketch the image of γ under the mapping z 7→ z - 3 i . (c) Sketch the image of γ under the mapping z 7→ 2 + 1 z - 3 i . Solution: Upper half of circle of radius 2 centered at 3 i ; upper half of circle of radius 2 centered at 0; lower half of circle of radius 1 2 centered at 2. 2. [3 marks] Find the points where the function f ( x + iy ) = x 2 ( i - 1) - 2 xy is differentiable, and find the derivative f 0 in terms of x and y at those points. Solution: Let u ( x, y ) = - x 2 - 2 xy and v ( x, y ) = x 2 be the real and imaginary parts of f , respectively. Then the second Cauchy-Riemann equation u y = - v x = - 2 x is identically satisfied. The first Cauchy-Riemann equation u x = v y is - 2( x + y ) = 0, which is satis- fied on the line y = - x . On this line, the derivative is given by f 0 ( x + iy ) = u x + iv x = 2 xi . 3. [3 marks] Let f ( z ) = p.v.( iz ) 1 2 . (a) Find f ( - 2) in a + bi form. (b) Where is f ( z ) analytic? Solution: We have f ( z ) = exp( 1 2 Log( iz )), so f ( - 2) = exp 1 2 Log( i ( - 2)) = exp 1 2 ln 2 - πi 4 = 1 - i. The function f ( z ) is analytic where Log( iz ) is, which is when iz does not lie on the non- positive real axis, or when z is not on the non-negative imaginary axis.
4. [3 marks] (a) Show that if | z | = 2, then | 3 z 2 - 4 z + 1 | ≥ 3. (b) Use the Estimation Lemma and the result in the previous part to show that I | z | =2 iz 2 + 5 3 z 2 - 4 z + 1 dz 12 π. Solution: If | z | = 2, then by the circle inequality, we have | 3 z 2 - 4 z + 1 | ≥ | 3 z 2 | - | - 4 z + 1 | = 12 - | - 4 z + 1 | ≥ 12 - 9 = 3 , since by the triangle inequality | - 4 z + 1 | ≤ | - 4 z | + 1 = 9. Then by the Estimation Lemma, we have I | z | =2 iz 2 + 5 3 z 2 - 4 z + 1 dz M · L, where M is any bound of iz 2 + 5 3 z 2 - 4 z + 1 on the contour and L = 4 π is the length of the contour. By the triangle inequality, we have | iz 2 + 5 | ≤ | iz 2 | + 5 = 9 on the coutour, and by the previous part | 3 z 2 - 4 z + 1 | ≥ 3. Therefore we can take M = 9 3 = 3, and then ML = 12 π . 5. [3 marks] (a) Evaluate H γ e i ( z - 2) dz , where γ is the circle | z | = 3, traversed anti-clockwise. (b) Evaluate R γ Log( z ) z dz , where γ is the semicircle | z | = 2 , Im( z ) 0, traversed anti- clockwise. Justify your answers. Solution: Since e iz is entire, the first integral is zero by the Cauchy-Goursat Theorem. For the second integral, we use the parametrization z = γ ( t ) = 2 e it , 0 t π , so that γ 0 ( t ) = iz . Then the integrand is ln 2 + it z · iz = - t + i ln 2 , so the integral is π Z 0 ( - t + i ln 2) dt = - π 2 2 + ln 2 .
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