Question

For the function \( \frac{\sin z}{z^4} \), the point z = 0 is

• A. A pole of order 4
• B. A pole of order 1
• C. A pole of order 2
• D. A pole of order 3

Hint:

Poles and Laurent Series. Expand the function in a Laurent series around the point \(z_0\). If the series has a finite number of terms with negative powers \((z-z_0)^{-k\), the singularity is a pole. The order of the pole is the largest positive integer \(k\) for which the coefficient of \((z-z_0)^{-k\) is non-zero. Alternatively, if \( \lim_{z\to z_0 (z-z_0)^m f(z) \) is finite and non-zero, then \(z_0\) is a pole of order m.

Answer 1

The Correct Option is D

Let \( f(z) = \frac{\sin z}{z^4} \)
We want to classify the singularity at \( z = 0 \)
We can use the Taylor series expansion for \( \sin z \) around \( z = 0 \): $$ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots $$ Substitute this into the function \( f(z) \): $$ f(z) = \frac{1}{z^4} \left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \right) $$ $$ f(z) = \frac{z}{z^4} - \frac{z^3}{3! z^4} + \frac{z^5}{5! z^4} - \dots $$ $$ f(z) = \frac{1}{z^3} - \frac{1}{3! z} + \frac{z}{5!} - \dots $$ This is the Laurent series expansion of \( f(z) \) around \( z = 0 \)
The principal part (terms with negative powers of \(z\)) contains terms up to \( \frac{1}{z^3} \)
The highest power of \( 1/z \) in the principal part determines the order of the pole
Since the highest power is 3 (from the term \( 1/z^3 \)), the function \( f(z) \) has a pole of order 3 at \( z = 0 \)