Question

The coefficient of \(\dfrac{1}{z}\) in the Laurent's series expansion of the function \[ f(z)=\frac{1}{z^2(1-z)} \] about \(z=0\), is ____.

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(2\)

Hint:

Remember the standard expansion: \[ \boxed{ \frac1{1-z} = 1+z+z^2+\cdots \qquad(|z|<1) } \] The coefficient of \(\dfrac1z\) in a Laurent series is called the residue at that point.

Answer 1

The Correct Option is C

Concept: For \[ |z|<1, \] the geometric series expansion is \[ \boxed{ \frac{1}{1-z} = 1+z+z^2+z^3+\cdots } \] Multiplying by \(\dfrac{1}{z^2}\) gives the Laurent series.

Step 1: Expand the function. \[ f(z) = \frac{1}{z^2} \left( 1+z+z^2+z^3+\cdots \right) \] \[ = \frac1{z^2} +\frac1z +1 +z +z^2+\cdots \]

Step 2: Identify the coefficient of \(\dfrac1z\). The coefficient of \[ \frac1z \] is \[ \boxed{1.} \] Therefore, the correct option is \[ \boxed{(C)\;1.} \]