Question:

The Laurent series expansion of f(z)=[2(z1)(z2)]f(z)=[\frac{2}{(z-1)(z-2)}] valid for z1>1|z-1|>1

Updated On: Mar 12, 2025
  • 1(z1)+2(z1)3+3(z1)5+.....\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^5}+.....
  • 1(z1)+2(z1)3+3(z1)3+.....\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....
  • 1(z2)+2(z2)3+3(z2)3+.....\frac{1}{(z-2)}+\frac{2}{(z-2)^3}+\frac{3}{(z-2)^3}+.....
  • 1(z1)+2(z2)3+3(z1)3+4(z2)4+.....\frac{1}{(z-1)}+\frac{2}{(z-2)^3}+\frac{3}{(z-1)^3}+\frac{4}{(z-2)^4}+.....
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

The correct option is (B): 1(z1)+2(z1)3+3(z1)3+.....\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....
Was this answer helpful?
0
0