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 $a_1/a_2=$ ____.

Question

Match List-I with List-II and choose the correct option:

LIST-I (Function)	LIST-II (Value)
(A) $ \int_{\gamma} \frac{1}{z-a} \, dz $, where $ \gamma: \|z-a\|=r, r > 0 $	(III) $ 2i\pi $
(B) $ \int_{\gamma} \frac{z+2}{z} \, dz $, where $ \gamma: z = 2e^{it}, 0 \le t \le \pi $	(IV) $ i\pi $
(C) $ \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz $, where $ \gamma: \|z\|=3 $	(II) $ 2i\pi(e^4 - e^2) $
(D) $ \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz $, where $ \gamma: \|z\|=2 $	(I) $ -4 + 2i\pi $

Choose the correct answer from the options given below:

Answer 1

Step 1: Since $(2z^2-13z+15)f(z)=1$ identically, substitute the Laurent series $f(z)=\sum a_nz^n$ and match coefficients of $z^{k}$ on both sides. For every integer $k\ge2$ (so that $k-1,k-2\ge0$), the right side contributes $0$, giving the recurrence $2a_{k-2}-13a_{k-1}+15a_k=0$, i.e. $a_k=\dfrac{13a_{k-1}-2a_{k-2}}{15}$.
Step 2: Find the roots of the denominator: $2z^2-13z+15=2(z-5)(z-3/2)$, and use the cover-up rule for partial fractions: $f(z)=\dfrac{1/7}{z-5}-\dfrac{1/7}{z-3/2}$ (coefficient at $z=5$ is $\frac{1}{2(5-3/2)}=\frac17$, and at $z=3/2$ is $\frac{1}{2(3/2-5)}=-\frac17$).
Step 3: Compute the first two nonnegative-index coefficients directly: expanding only the $\dfrac{1/7}{z-5}=-\dfrac{1}{35}\left(1+\dfrac{z}{5}+\cdots\right)$ term (the $\dfrac{1/7}{z-3/2}$ term contributes only negative powers of $z$ in this annulus) gives $a_0=-\dfrac{1}{35}$ and $a_1=-\dfrac{1}{175}$.
Step 4: Apply the recurrence from Step 1 with $k=2$: $a_2=\dfrac{13a_1-2a_0}{15}=\dfrac{13\left(-\frac{1}{175}\right)-2\left(-\frac{1}{35}\right)}{15}=\dfrac{-\frac{13}{175}+\frac{10}{175}}{15}=\dfrac{-\frac{3}{175}}{15}=-\dfrac{1}{875}$.
Step 5: This confirms $a_1=-\dfrac{1}{175}$, $a_2=-\dfrac{1}{875}$, so\[\dfrac{a_1}{a_2}=\dfrac{-1/175}{-1/875}=5.\]\[\boxed{\dfrac{a_1}{a_2}=5}\]

LIST-I (Function)	LIST-II (Value)
(A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: \|z-a\|=r, r > 0 \)	(III) \( 2i\pi \)
(B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \)	(IV) \( i\pi \)
(C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: \|z\|=3 \)	(II) \( 2i\pi(e^4 - e^2) \)
(D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: \|z\|=2 \)	(I) \( -4 + 2i\pi \)

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 \(a_1/a_2=\) ____.

Show Hint

The Correct Option is D

Solution and Explanation

Top Questions on Complex Analysis

Questions Asked in CPET exam