GATE 2022 IN Q.50: solution and answer

GATE 2022 · INQ.50 · 2 marks · MSQ

For the complex number $z = \frac{a + j b}{a - j b}$ , where $a > 0$ and $b > 0$ .

Which of the following statement(s) is/are true?

(A)
The phase is $2 tan^{- 1} \frac{b}{a}$
(B)
The phase is $tan^{- 1} \frac{2 b}{a}$
(C)
The magnitude is $1$
(D)
The magnitude is $\frac{a ^{2} + b ^{2}}{a ^{2} - b ^{2}}$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

(A) and (C) are true. $z = e^{j 2 t a n^{- 1} (b / a)}$ , so its magnitude is $1$ and its phase is $2 tan^{- 1} \frac{b}{a}$ .

Official answer: A, CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Write the numerator in polar form: $a + j b = r e^{j θ}$ with $r = a^{2} + b^{2}$ and $θ = tan^{- 1} \frac{b}{a}$ . Since $a > 0$ and $b > 0$ , the point lies in the first quadrant, so $θ \in (0, \frac{π}{2})$ .
The denominator is the complex conjugate of the numerator: $a - j b = \overline{a + j b} = r e^{- j θ}$ .
Divide in polar form (magnitudes divide, angles subtract): $z = \frac{r e ^{j θ}}{r e ^{- j θ}} = e^{j 2 θ}$
Magnitude: $∣ z ∣ = \frac{∣ a + j b ∣}{∣ a - j b ∣} = \frac{a ^{2} + b ^{2}}{a ^{2} + b ^{2}} = 1$ . So (C) is true and (D) is false. (D) would need $∣ a - j b ∣ = a^{2} - b^{2}$ , but the modulus of $a - j b$ is $a^{2} + (- b)^{2} = a^{2} + b^{2}$ .
Phase: $∠ z = θ - (- θ) = 2 θ = 2 tan^{- 1} \frac{b}{a}$ . Because $θ \in (0, \frac{π}{2})$ , we get $2 θ \in (0, π)$ , which is already a principal argument, so no multiple of $2 π$ has to be added or removed. So (A) is true.
Check (B). For $a \neq = b$ , the double-angle formula $tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ gives $tan (∠ z) = \frac{2 ( b / a )}{1 - ( b / a ) ^{2}} = \frac{2 ab}{a ^{2} - b ^{2}}$ If (B) were true, we would have $tan (∠ z) = \frac{2 b}{a}$ . Setting $\frac{2 ab}{a ^{2} - b ^{2}} = \frac{2 b}{a}$ gives $a^{2} = a^{2} - b^{2}$ , so $b = 0$ , which is excluded. For $a = b$ , $∠ z = 2 \cdot \frac{π}{4} = \frac{π}{2}$ , while $tan^{- 1} 2 \approx 63. 4^{\circ}$ . So (B) is false for every $a > 0, b > 0$ .
Correct options: (A) and (C).

KEY IDEA

$a - j b$ is the conjugate of $a + j b$ . The two have the same modulus and opposite angles. Dividing a complex number by its conjugate therefore gives a number of magnitude $1$ whose angle is twice the original angle: $\frac{r e ^{j θ}}{r e ^{- j θ}} = e^{j 2 θ}$ .

WATCH OUT FOR

Doubling inside the inverse tangent, writing $tan^{- 1} \frac{2 b}{a}$ instead of $2 tan^{- 1} \frac{b}{a}$ (option B). The inverse tangent is not linear, and $tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ , not $2 tan θ$ . A second trap: rationalising to $z = \frac{( a ^{2} - b ^{2} ) + j 2 ab}{a ^{2} + b ^{2}}$ and then taking the phase as $tan^{- 1} \frac{2 ab}{a ^{2} - b ^{2}}$ gives the wrong quadrant when $b > a$ , because the real part is then negative. A third: taking $∣ a - j b ∣ = a^{2} - b^{2}$ (option D). The minus sign is inside the complex number, and the modulus is still $a^{2} + b^{2}$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 24 Complex analysis questions, with notes

GATE 2022 IN Q.50: solution and answer

Answer and solution

More Complex analysis questions