By William R. Derrick

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E x a m p l e 1 Since cos ζ is entire, has antiderivative sin z, and ^ is simply connected, we have cos ζ dz = sin ζ = 2 sin i = 2i sinh(l). and along any pwd closed curve γ cos ζ dz = 0. E x a m p l e 2 The function l/z, analytic in ζΦΟ, has antiderivative log ζ. In this case care must be taken to specify the domain G. Suppose G is given by I arg ζ | < π, then for any arc joining — / to / in G ' dz ^ — = Log ζ -i ζ = πι. On the other hand, if G' is the domain given by 0 < arg ζ < 2π, we have dz — = log ζ Finally, letting y: ζ{θ) = e'\ π/2 < Ö < 3π/2, we obtain dz |z| = l 3π ζ -i r in Ζ = r dz -π/2<ο<π/2, and γ': ζ(θ) = e'", r dz — + ^ r —ζ = πι Jy ζ since y lies in G and y' lies in G\ Example 3 | r sin(x") ^ π dx = - .

Sin ζ z^ + 1 (b) dz. (d) ' cos ζ dz. sin ζ ζ — ζ ί/ζ. 2 . 4 T H E C A U C YH I N T E G R A L 53 F O R M U L A 3. Let y: z(t) = 2e'^ 4- 1, 0 < ί < 2π. Evaluate the following integrals: (a) dz, cos ζ (b) (z-1) 'y (c) sin ζ sin ζ dz, id) dz, 2 dz. 4. Let / ( ζ ) be analytic in |z - C| < ^ . Prove Gauss's Mean-Value /(() = -L r'7(C 2π -^0 + re'') de, Theorem 0

Find all the zeros of sinh ζ and cosh z. 10. Show that the function vv = sin ζ maps (a) (b) (c) the strip | A : | < π/2 onto ^ - {z: j = 0, \x\ > 1}, the semiinfinite strip \x\ < n/l, y>0 onto the upper half plane, the semiinfinite strip 0 < χ < π/2, y >0 onto the first quadrant, by indicating what happens to horizontal and vertical line segments under the transformation w = sin ζ = sin χ cosh y + i cos χ sinh y. NOTES Section LI Formulas relating ζ to Ζ in the stereographic projection are easy to compute: [A, pp.