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

Concept: To find the Laurent (or Taylor) series expansion of a function within a specified disk or annulus, we use the standard geometric series expansion: \[ \frac{1}{1+t} = (1+t)^{-1} = \sum_{n=0}^{\infty} (-1)^n t^n \quad \text{valid for } |t| \lt 1 \] The given convergence region is the open disk \( |z| \lt 2 \). This convergence constraint means that:

• For terms with a denominator like \( z+2 \), we must factor out 2 so that the variable term becomes \( \frac{z}{2} \), since \( |z| \lt 2 \implies \left|\frac{z}{2}\right| \lt 1 \).

• For terms with a denominator like \( z+3 \), since \( |z| \lt 2 \lt 3 \), it follows that \( |z| \lt 3 \implies \left|\frac{z}{3}\right| \lt 1 \). Thus, we factor out 3.

Step 1: Setting up the expansion for the second term \( \frac{3}{z+2} \).
Let us rewrite this fraction to set up our geometric series form by factoring out the constant 2 from the denominator: \[ \frac{3}{z+2} = \frac{3}{2\left(1 + \frac{z}{2}\right)} = \frac{3}{2} \left( 1 + \frac{z}{2} \right)^{-1} \] Since our region specifies \( |z| \lt 2 \), the absolute value satisfies \( \left|\frac{z}{2}\right| \lt 1 \). Applying the binomial series expansion: \[ \frac{3}{2} \left( 1 + \frac{z}{2} \right)^{-1} = \frac{3}{2} \sum_{n=0}^{\infty} (-1)^n \left(\frac{z}{2}\right)^n \quad \cdots (1) \]

Step 2: Setting up the expansion for the third term \( \frac{8}{z+3} \).
We use the same approach for the next fraction, factoring out the constant 3 from the denominator: \[ \frac{8}{z+3} = \frac{8}{3\left(1 + \frac{z}{3}\right)} = \frac{8}{3} \left( 1 + \frac{z}{3} \right)^{-1} \] Since \( |z| \lt 2 \), it is also true that \( |z| \lt 3 \), which means \( \left|\frac{z}{3}\right| \lt \frac{2}{3} \lt 1 \). This justifies expanding it as a geometric series: \[ \frac{8}{3} \left( 1 + \frac{z}{3} \right)^{-1} = \frac{8}{3} \sum_{n=0}^{\infty} (-1)^n \left(\frac{z}{3}\right)^n \quad \cdots (2) \]

Step 3: Combining all the terms to form the full series expansion.
The complete function given in the problem is: \[ f(z) = 1 + \frac{3}{z+2} - \frac{8}{z+3} \] Substituting our two derived series expressions (1) and (2) directly back into this equation yields: \[ f(z) = 1 + \frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{2}\right)^n - \frac{8}{3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n \] This matches Option (C).