Question

If \(\displaystyle\sum_{n=-\infty}^{\infty}a_nz^n\) is the Laurent series expansion of the function \(f(z)=\dfrac{1}{2z^2-13z+15}\) in the annulus \(\{z\in\mathbb{C}:3/2<|z|<5\}\), then \(a_1/a_2=\) ____.

• A. \(-5\)
• B. \(-\dfrac{1}{5}\)
• C. \(\dfrac{1}{5}\)
• D. \(5\)

Hint:

Split into partial fractions at the roots \(5\) and \(3/2\), then expand each part in the direction valid for the given annulus.

Answer 1

The Correct Option is D

Step 1: Solve \(2z^2-13z+15=0\) using the quadratic formula: \(z=\dfrac{13\pm\sqrt{169-120}}{4}=\dfrac{13\pm7}{4}\), giving \(z=5\) and \(z=3/2\). So \(2z^2-13z+15=2(z-5)(z-3/2)\).

Step 2: Decompose into partial fractions: \(\dfrac{1}{(z-5)(z-3/2)}=\dfrac{1}{5-3/2}\left(\dfrac{1}{z-5}-\dfrac{1}{z-3/2}\right)=\dfrac{2}{7}\left(\dfrac{1}{z-5}-\dfrac{1}{z-3/2}\right)\). Dividing by the extra factor \(2\): \(f(z)=\dfrac{1}{7}\left(\dfrac{1}{z-5}-\dfrac{1}{z-3/2}\right)\).

Step 3: In the annulus \(3/2<|z|<5\), expand \(\dfrac{1}{z-5}\) in powers of \(z\) (valid since \(|z|<5\)): \(\dfrac{1}{z-5}=-\dfrac{1}{5}\cdot\dfrac{1}{1-z/5}=-\sum_{n=0}^{\infty}\dfrac{z^n}{5^{n+1}}\).

Step 4: Expand \(\dfrac{1}{z-3/2}\) in powers of \(1/z\) (valid since \(|z|>3/2\)): \(\dfrac{1}{z-3/2}=\dfrac{1}{z}\cdot\dfrac{1}{1-\frac{3}{2z}}=\sum_{n=0}^{\infty}\dfrac{(3/2)^n}{z^{n+1}}\), which contributes only to negative powers of \(z\).

Step 5: So for \(n\ge0\), \(a_n=-\dfrac{1}{7\cdot5^{n+1}}\). Thus \(a_1=-\dfrac{1}{7\cdot5^2}=-\dfrac{1}{175}\) and \(a_2=-\dfrac{1}{7\cdot5^3}=-\dfrac{1}{875}\).

Step 6: Therefore \(\dfrac{a_1}{a_2}=\dfrac{-1/175}{-1/875}=\dfrac{875}{175}=5\).

\[\boxed{\dfrac{a_1}{a_2}=5}\]