Question

Which of the following functions is/are expandable using Maclaurin series?

• A. $\ln(1+z)$
• B. $\ln z$
• C. $\dfrac{1}{z^2}$
• D. $\exp(z)$

Hint:

For Maclaurin expansions, always check if the function is analytic at \( z = 0 \).

If the function has a singularity (like \( \ln z \) or \( \tfrac{1}{z^2} \)), it cannot be expanded.

Entire functions like \( e^z \), \( \sin z \), \( \cos z \), and power series like \( \ln(1+z) \) (valid for \( |z| < 1 \)) are always expandable.

Answer 1

The Correct Option is A, D

Step 1: Recall the definition of a Maclaurin series.

A Maclaurin series is simply the Taylor series expansion of a function \( f(z) \) around \( z = 0 \): \( f(z) = f(0) + f'(0)\tfrac{z}{1!} + f''(0)\tfrac{z^2}{2!} + f^{(3)}(0)\tfrac{z^3}{3!} + \cdots \). This requires \( f(z) \) to be analytic (differentiable in a neighborhood of \( z=0 \)).

Step 2: Check each option.

• (A) \( \ln(1+z) \): analytic at \( z=0 \). Expansion: \( \ln(1+z) = z - \tfrac{z^2}{2} + \tfrac{z^3}{3} - \tfrac{z^4}{4} + \cdots \), valid for \( |z| < 1 \). Hence, expandable.
• (B) \( \ln z \): undefined at \( z=0 \) (diverges). Not expandable.
• (C) \( \tfrac{1}{z^2} \): has a pole at \( z=0 \). Not expandable.
• (D) \( e^z \): entire function. Expansion: \( e^z = 1 + z + \tfrac{z^2}{2!} + \tfrac{z^3}{3!} + \cdots \). Hence, expandable.

Step 3: Final conclusion.

Only options (A) \( \ln(1+z) \) and (D) \( e^z \) are expandable using the Maclaurin series. Answer: \( \boxed{\text{(A) and (D)}} \).