Pleae help with a The line segments from 20 to 21 to 21 to 2

Pleae help with:

(a) The line segments from (-2,0) to (-2,-1) to (2,-1) to (2, 0)

(b) the lower half of a circle

(c) the upper half of a circle.

Thank you!

Solution

1.

Let f(t) = eit, and t are real. We obtain Z 2 0 eit dt Z 2 0 |eit| dt = 2. The left-hand side of the above inequality is equal to Z 2 0 eit dt = eit i 20 = |e2i 1| || . Combining the results, we obtain |e2i 1| 2||, is real. 4 Definition of a contour integral Consider a curve C which is a set of points z = (x, y) in the complex plane defined by x = x ( t ), y = y ( t ), a t b, where x ( t) and y ( t) are continuous functions of the real parameter t. One may write z ( t) = x ( t) + iy ( t ), a t b. • The curve is said to be smooth if z ( t) has continuous derivative z ( t ) 6= 0 for all points along the curve. • A contour is defined as a curve consisting of a finite number of smooth curves joined end to end. A contour is said to be a simple closed contour if the initial and final values of z ( t) are the same and the contour does not cross itself. 5 • Let f ( z) be any complex function defined in a domain D in the complex plane and let C be any contour contained in D with initial point z 0 and terminal point z. • We divide the contour C into n subarcs by discrete points z 0, z 1, z 2, . . ., z n 1, z n = z arranged consecutively along the direction of increasing t. • Let k be an arbitrary point in the subarc z k z k+1 and form the sum nX 1 k=0 f ( k)( z k+1 z k ).

, C = C1 + C2 + C3 + C4, and a parameterization for the curves is C1 : z(t) = t, 0 t 1 = z (t) = 1 C2 : z(t) = 1 + (t 1)i, 1 t 2 = z (t) = i C3 : z(t) = (3 t) + i, 2 t 3 = z (t) = 1 C4 : z(t) = (4 t)i, 3 t 4 = z (t) = i 2 Integrate f(z) along each contour. C1 f(z) dz = 1 0 f[z(t)]z (t) dt = 1 0 et · 1 dt = e t 1 0 = e 1. C2 f(z) dz = 2 1 f[z(t)]z (t) dt = 2 1 exp((1 (t 1)i))i dt = 2 1 ie e (t1)i dt = e 2 1 ie(t1)i dt = e ( e (t1)i 2 1 ) = e ( 1 e i) = 2e . C3 f(z) dz = 3 2 f[z(t)]z (t) dt = 3 2 exp(((3 t) i))(1) dt = 3 2 eie (3t) dt = e i e (3t) 3 2 = 1 (1 e ) = e 1. C4 f(z) dz = 4 3 f[z(t)]z (t) dt = 4 3 exp((4 t)i)(i) dt = 4 3 ie(t4)i