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

Question

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

cdquestions Admin · Accepted Answer

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

Answer

$\frac{1}{(z-1)}+\frac{2}{(z-1)^3}+\frac{3}{(z-1)^5}+.....$

Answer

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

Answer

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

Answer

$\frac{1}{(z-1)}+\frac{2}{(z-2)^3}+\frac{3}{(z-1)^3}+\frac{4}{(z-2)^4}+.....$

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