- Fn Z 1 Zn Ez 1 N 0 1 2 E Let Y Be The Keyhole Contoury Shown Below With Inner Radius R 1 And Out 1 (4.08 KiB) Viewed 31 times
- Fn Z 1 Zn Ez 1 N 0 1 2 E Let Y Be The Keyhole Contoury Shown Below With Inner Radius R 1 And Out 2 (71.48 KiB) Viewed 31 times
e) Let y be the keyhole contoury shown below, with inner radius r = 1, and outer radius R = (2N+1)π, NE N. Evaluate f fn(z)dz. [5] f) Take the limit N → ∞,6 → 0 to show that, for n > 1, [5] ∞ 1 1 Bn = Σ -(n!) (2πi)n mn (-m)n + m=1 so if n ≥ 1, B2n+1 = 0 and B2n is purely real and proportional to $(2n). Hint: In the circle |z| = (2N + 1)π, |e² − 1| > 1/2. R=(2N+1) 2лNi 2лi -2πi -2лNi r=1 R+i8