GATE 2023 IN Q.19: solution and answer

GATE 2023 · INQ.19 · 1 mark · MCQ

$F (z) = \frac{1}{1 - z}$ , when expanded as a power series around $z = 2$ , would result in $F (z) = \sum_{k = 0}^{\infty} a_{k} (z - 2)^{k}$ , with the region of convergence (ROC) $∣ z - 2∣ < 1$ . The coefficients $a_{k}$ , $k \geq 0$ , are given by the expression ______.

(A)
$(- 1)^{k}$
(B)
$(- 1)^{k + 1}$
(C)
$(\frac{1}{2})^{k}$
(D)
$(\frac{- 1}{2})^{k + 1}$

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Answer and solution

ANSWER

$a_{k} = (- 1)^{k + 1}$ , so $F (z) = - 1 + (z - 2) - (z - 2)^{2} + \dots$ for $∣ z - 2∣ < 1$ . Option (B).

Official answer: BSource: Official answer key, GATE 2023 IN (IIT Kanpur)

Substitute $w = z - 2$ , so $z = w + 2$ and the ROC becomes $∣ w ∣ < 1$ . Then $1 - z = 1 - (w + 2) = - (1 + w)$ .
So $F (z) = \frac{1}{1 - z} = - \frac{1}{1 + w} .$
Use the geometric series $\frac{1}{1 + w} = \sum_{k = 0}^{\infty} (- 1)^{k} w^{k}$ . It holds for $∣ w ∣ < 1$ , which is exactly the given ROC $∣ z - 2∣ < 1$ .
Hence $F (z) = - k = 0 \sum \infty (- 1)^{k} (z - 2)^{k} = k = 0 \sum \infty (- 1)^{k + 1} (z - 2)^{k},$ so $a_{k} = (- 1)^{k + 1}$ .
Check with Taylor's formula $a_{k} = \frac{F ^{(k)} ( 2 )}{k !}$ . Since $F^{(k)} (z) = \frac{k !}{( 1 - z ) ^{k + 1}}$ , we get $a_{k} = \frac{1}{( 1 - 2 ) ^{k + 1}} = (- 1)^{k + 1}$ . As a quick test, $a_{0} = F (2) = \frac{1}{1 - 2} = - 1$ . Option (A) gives $a_{0} = 1$ , (C) gives $1$ and (D) gives $- \frac{1}{2}$ , so only (B) fits. Answer: option (B).

KEY IDEA

Shift to the new centre with $w = z - 2$ and rewrite $1 - z$ as $- (1 + w)$ . That turns $F$ into $- \frac{1}{1 + w}$ , a geometric series that converges for $∣ w ∣ < 1$ . You can confirm the result in one line with $a_{0} = F (2) = - 1$ .

WATCH OUT FOR

The usual slip is dropping the minus sign in $1 - z = - (1 + w)$ . That gives $a_{k} = (- 1)^{k}$ , which is option (A). A second trap is mixing up the centre: the familiar $\frac{1}{1 - z} = \sum z^{k}$ is centred at $z = 0$ , and forcing a factor of $\frac{1}{2}$ into the expansion leads to options (C) or (D), which belong to a different expansion with a different ROC. Checking $a_{0} = F (2) = - 1$ catches all three.

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GATE 2023 IN Q.19: solution and answer

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