GATE 2023 EC Q.13: solution and answer

GATE 2023 · ECQ.13 · 1 mark · MCQ

Let $w^{4} = 16 j$ . Which of the following cannot be a value of $w$ ?

(A)
$2 e^{\frac{j 2 π}{8}}$
(B)
$2 e^{\frac{j π}{8}}$
(C)
$2 e^{\frac{j 5 π}{8}}$
(D)
$2 e^{\frac{j 9 π}{8}}$

See the question in the paper

Answer and solution

ANSWER

(A) $2 e^{j 2 π /8}$ cannot be a value of $w$ . Its fourth power is $16 e^{j π} = - 16$ , not $16 j$ . The four fourth roots of $16 j$ are $2 e^{j π /8}$ , $2 e^{j 5 π /8}$ , $2 e^{j 9 π /8}$ and $2 e^{j 13 π /8}$ , and options (B), (C) and (D) are three of them.

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

Write the right-hand side in polar form: $16 j = 16 e^{j π /2}$ . The argument is fixed only up to multiples of $2 π$ , so in general $16 j = 16 e^{j (π /2 + 2 k π)}$ for any integer $k$ .
Take fourth roots (De Moivre's theorem for $n$ -th roots): $w = 1 6^{1/4} e^{j (\frac{π /2 + 2 k π}{4})} = 2 e^{j (\frac{π}{8} + \frac{k π}{2})}, k = 0, 1, 2, 3$
List the four roots. They lie on the circle $∣ w ∣ = 2$ , spaced $π /2 = 4 π /8$ apart: $w \in {2 e^{j π /8}, 2 e^{j 5 π /8}, 2 e^{j 9 π /8}, 2 e^{j 13 π /8}}$
Match the options: (B) $2 e^{j π /8}$ is $k = 0$ , (C) $2 e^{j 5 π /8}$ is $k = 1$ and (D) $2 e^{j 9 π /8}$ is $k = 2$ . The argument $2 π /8$ in (A) differs from $π /8$ by $π /8$ , which is not a multiple of $π /2$ , so (A) is not one of the roots.
Check (A) directly: $(2 e^{j 2 π /8})^{4} = 16 e^{j 8 π /8} = 16 e^{j π} = - 16 \neq = 16 j$ . So (A) cannot be a value of $w$ .

KEY IDEA

The $n$ -th roots of $r e^{j θ}$ are $r^{1/ n} e^{j (θ + 2 k π) / n}$ for $k = 0, 1, \dots, n - 1$ . They sit on a circle of radius $r^{1/ n}$ , spaced $2 π / n$ apart. For $16 j$ that means radius $2$ , starting angle $π /8$ and spacing $π /2$ . Any option whose angle is not $π /8$ plus a multiple of $π /2$ is not a root, and raising it to the fourth power confirms this.

WATCH OUT FOR

A common mistake is to take the argument of $16 j$ as $π /4$ instead of $π /2$ , or to divide only the principal argument by 4 and forget the $2 k π /4 = k π /2$ spacing. Either slip gives the wrong set of roots, and then (A), with angle $2 π /8 = π /4$ , looks like a valid root. Another slip is to read "cannot be" as "can be" and pick (B), the principal root.

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GATE 2023 EC Q.13: solution and answer

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