Question

The Laurent's series for the function \( f(z) = 1 + \frac{3}{z+2} - \frac{8}{z+3} \) in the region \( |z| \lt 2 \) is

• A. \( \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n \)
• B. \( 1 + \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n \)
• C. \( 1 + \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n \)
• D. \( \frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^n(-1)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n \)

Hint:

Always look closely at the convergence condition! To ensure your geometric series converges (\( |t| \lt 1 \)), always factor out the larger value in magnitude between the variable \( z \) and the constant term.

Answer 1

The Correct Option is C

Step 1: Locate the singularities and check the radius each term allows.
The function \( f(z) = 1 + \dfrac{3}{z+2} - \dfrac{8}{z+3} \) has poles at \( z = -2 \) and \( z = -3 \), at distances \( 2 \) and \( 3 \) from the origin. We want the expansion valid on \( |z| \lt 2 \), which is smaller than both distances, so both fractional terms can be written as ordinary geometric series in powers of \( z \).

Step 2: Expand the term with the closer pole, \( \dfrac{3}{z+2} \).
Write it with the pole's distance pulled out: \[ \frac{3}{z+2} = \frac{3}{2}\cdot\frac{1}{1+\frac{z}{2}} \] Since \( |z| \lt 2 \) means \( \left|\frac{z}{2}\right| \lt 1 \), the geometric series \( \frac{1}{1+t} = \sum(-1)^n t^n \) applies with \( t = z/2 \): \[ \frac{3}{z+2} = \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n \]
Step 3: Expand the term with the farther pole, \( \dfrac{8}{z+3} \), the same way.
Pull out its own pole distance: \[ \frac{8}{z+3} = \frac{8}{3}\cdot\frac{1}{1+\frac{z}{3}} = \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n, \quad \text{valid for } \left|\frac{z}{3}\right| \lt 1 \] Since our region \( |z| \lt 2 \) sits well inside \( |z| \lt 3 \), this expansion holds throughout.

Putting both series back with the constant term gives \[ f(z) = 1 + \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n \] which is option (C). Each term keeps its own pole's distance, 2 for the first sum and 3 for the second, since that is what fixes the radius of convergence for each piece.