Returning to Eq. (1), we now impose the boundary condition u ( r = a, θ ) = u 0 , and solve for the coefficients by inverting the Fourier sine series; we do the inversion in the usual way, by multiplying both sides by a trig function with index m , integrating over angles, and using orthogonality on the RHS to collapse the sum to n = m ˆ β 0 u ( r = a, θ ) sin mπ β θ dθ = X n B n a k n ˆ β 0 sin nπ β θ sin mπ β θ dθ = B m a k m × β 2 The LHS in the above equation works out as follows LHS = u 0 ˆ β 0 sin mπ β θ dθ = u 0 β mπ - cos mπ β θ β 0 = 2 u 0 β mπ , m = odd = 0 , m = even So putting together the last two equations, we at last have an expression for the coefficients B n a k n = 4 u 0 nπ , n = odd
4 Problem II (a) Use the Cauchy-Riemann equations to show that the function sin z is analytic through- out the finite complex plane. (b) Show that if a function f ( z ) is analytic in a region | z | < R, then the function f (1 /z ) is analytic in the region | z | > 1 /R . (c) Find the Laurent series expansion for the function f ( z ) = z 2 sin 1 z (d) Use the residue theorem to evaluate the integral I = ‰ C z 2 sin 1 z dz where C is a counter-clockwise oriented closed contour that encloses the origin. Solution (a) We can use the standard sine identity to find the real and imaginary parts of the function f ( z ) = sin ( x + iy ) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y = u ( x, y ) + iv ( x, y ) So now let’s check that the Cauchy-Riemann equations are satisfied. First, ∂u ∂x = cos x cosh y ? = ∂v ∂y = cos x cosh y which is true throughout the finite complex plane (the point at infinity is excluded, since the hyperbolic trig functions blow up). Second ∂u ∂y = cos x sinh y ? = - ∂v ∂x = - cos x sinh y which is also true throughout the finite complex plane. Solution (b) We want to prove that the expression d dz f 1 z exists in the region | z | > 1 /R . This is almost trivial, given that f ( z ) is analytic in the region | z | < R . However, a complete proof requires a little extra finesse. To do this carefully, define w = 1 z
5 and use the chain rule to write d dz f 1 z = dw dz × df ( w ) dw = - w 2 df ( w ) dw Now the derivative on the RHS exists for | w | < R , and since w 2 is finite in this domain, the derivative on the LHS also exists in this domain. Thus we have established that f (1 /z ) is analytic for | z | > 1 /R . Solution (c) Thanks to the previous proof, we can simply use the Taylor series expansion for sin z to work this out z 2 sin 1 z = z 2 ∞ X n =0 1 (2 n + 1)! 1 z 2 n +1 = z + ∞ X n =1 c - n z 1 - 2 n where I split off the regular and principal parts of the series, with c - n = 1 / (2 n + 1)! Note about convergence: since sin z is analytic for all finite z , sin(1 /z ) is anaytical for all z > 0 .
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