MAA 4402 Lecture 9: Chapter 9 Notes

• School: UCF
• Department: Mathematics - Analysis
• Course: MAA 4402
• Professor: Bhimsen Shivamoggi

Document Summary

2z2 z+1 (z 1)2(z+1) = a z+1 (z 1)2 z 1 + c (z 1)2(z+1) = a(z+1)(z 1)2 (z 1)2 + b. Then (z + 1)(z 1)2 2z2 z+1 implies 2z2 z + 1 = a(z + 1) + b(z + 1)(z 1) + c(z 1)2. Plug in 1 we get 2 1 + 1 = a(1 + 1) + 0 + 0 so a = 1. Plug in -1 we get 2 + 1 + 1 = 0 + 0 + 4c so c = 1. Plug in 0 we get 1 = a(1) + b( 1)( 1) + c( 1)2 so 1 = a b + c using what. A and c equals we get 1 = 1 b + 1 so b = 1 z+1 dz = r . 8i3 (z+i)2 (z i)2 dz+r|z|=3 eiz d (z+i)4 (z+i)2 )|z=i = ei2 (z2+1)2 = r|z|=3.