Download Complex Analysis with Applications - Midterm Exam with Solutions | MATH 417 and more Exams Mathematics in PDF only on Docsity! Math 417 – Midterm Exam Solutions Friday, July 11, 2008 1. (30 pts) Determine all possible series representations of the function: f(z) = 1 z(z2 + 1) about z = 0 and state their regions of validity. Solution: The function has singular points at z = 0, z = i, and z = −i. It will have a Laurent Series representation in the region 0 < |z| < 1 and another Laurent Series representation in the region 1 < |z| < ∞. I. In the region 0 < |z| < 1 we have f(z) = 1 z · 1 1 + z2 = 1 z ( 1 − z2 + (z2)2 − (z2)3 + · · · ) = 1 z − z + z3 − z5 + · · · = ∞ ∑ n=0 (−1)nz2n−1 II. In the region 1 < |z| < ∞ we have f(z) = 1 z · 1 1 + z2 = 1 z · 1 z2(1 + 1 z2 ) = 1 z3 · 1 1 + 1 z2 = 1 z3 ( 1 − 1 z2 + ( 1 z2 )2 − ( 1 z2 )3 + · · · ) = 1 z3 − 1 z5 + 1 z7 − 1 z9 = ∞ ∑ n=0 (−1)n z2n+3 2. (20 pts) Find all points at which f(z) = x x2 + y2 − i y x2 + y2 is differentiable. At what points is f analytic? Explain. 3. (40 pts) Compute each of the following integrals: (a) ∫ C sin ( 1 z ) dz where C is the circle |z| = 1 oriented counterclockwise. (b) ∫ C ez z(z2 + 1) dz where the contour C is shown below: x y i -i C Solution: (a) The function f(z) = sin(1 z ) has a singular point at z = 0, which lies inside C. The Laurent Series of f(z) around z = 0 in the region 0 < |z| < ∞ is f(z) = sin ( 1 z ) = 1 z − 1 3!z3 + 1 5!z5 − 1 7!z7 + · · · We can see that z = 0 is an essential singularity and that the residue at z = 0 is Res z=0 f(z) = c −1 = 1 Using the Cauchy Residue Theorem, the value of the integral is ∫ C sin ( 1 z ) dz = 2πi Res z=0 f(z) = 2πi (b) The function has singular points at z = 0, z = i, and z = −i. I. Let φ1(z) = ez z2+1 . Clearly, φ1(z) is analytic and nonzero at z = 0 and f(z) = φ1(z) z . Therefore, z = 0 is a simple pole and the residue is Res z=0 f(z) = φ1(0) = 1 II. Let φ2(z) = ez z(z+i) . Clearly, φ2(z) is analytic and nonzero at z = i and f(z) = φ2(z) z−i . Therefore, z = i is a simple pole and the residue is Res z=i f(z) = φ2(i) = − ei 2
