Excercise Sheet Solution 5

# Excercise Sheet Solution 5 - MATH 20142 Complex Analysis...

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MATH 20142 Complex Analysis Solution Sheet 5 1. (a) We have z ( t ) = e it = cos t - i sin t ( - π t 0), whence our curve is the semicircle (in the clockwise direction). (b) Here z ( t ) = x ( t ) + iy ( t ), where x ( t ) = 1 + 2 cos t, y ( t ) = 1 + 2 sin t, 0 t 2 π . Hence ( x - 1) 2 + ( y - 1) 2 = 4, i.e., we have the circle centred at (1 , 1) with radius 2 (in the counterclockwise direction). (c) z ( t ) = x ( t ) + iy ( t ), where x ( t ) = t, y ( t ) = exp t , whence y = exp x , i.e., we have a piece of the graph of exp from 0 to 1. 2. (a) Along γ , we have Re z 0, whence γ Re z dz = 0. (b) Put σ 1 ( t ) = t, 0 t 1. Here Re( σ 1 ( t )) = t, σ 1 ( t ) = 1, whence σ 1 Re z dz = 1 0 t dt = 1 2 . Put σ 2 ( t ) = 1 + ( i - 1) t, 0 t 1. We have Re σ 2 ( t ) = 1 - t, σ 2 ( t ) = - 1 + i , whence σ 2 Re z dz = ( - 1 + i ) 1 0 (1 - t ) dt = - 1 + i 2 . Finally, σ Re z dz = σ 1 Re z dz + σ 2 Re z dz = i 2 . The reason why the answers differ is that Re z has no antiderivative hence the FTCI does not apply. 3. The contour γ is the segment of the straight line from - i to i . The contour σ is a semicircle. Along γ , we have γ ( t ) = - i +2 it, 0 t 1, whence | γ ( t ) | = | 2 t - 1 | and γ ( t ) = 2 i . Thus, γ | z | dz = 2 i 1 0 | 2 t - 1 | dt = 2 i ( 1 / 2 0 (1 - 2 t ) dt + 1 1 / 2 (2 t - 1) dt ) = 4 i · 1 / 2 0 (1 - 2 t ) dt = i. 1

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Along σ , | z | ≡ 1, whence
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