Question

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

• A. $\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^5}+.....$
• B. $\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....$
• C. $\frac{1}{(z-2)}+\frac{2}{(z-2)^3}+\frac{3}{(z-2)^3}+.....$
• D. $\frac{1}{(z-1)}+\frac{2}{(z-2)^3}+\frac{3}{(z-1)^3}+\frac{4}{(z-2)^4}+.....$

Answer 1

Answer: (B)

$\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....$

Answer 2

The correct option is (B): $\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^3}+.....$