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JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Complex Numbers

Let

(- 2 - \frac{1}{3} i)^{3} = \frac{x + iy}{27} (i = - 1)

, where x and y are real numbers, then

y - x

equals:

(A)91✓ Correct
(B)-85
(C)85
(D)-91

Answer: (A)

Solution

Pull the fraction out first:

- 2 - \frac{i}{3} = - \frac{6 + i}{3}

, so

(- 2 - \frac{i}{3})^{3} = - \frac{( 6 + i ) ^{3}}{27}

, and comparing with

\frac{x + i y}{27}

gives

x + i y = - (6 + i)^{3}

. Now

(6 + i)^{2} = 35 + 12 i

, and

(6 + i)^{3} = (35 + 12 i) (6 + i) = 198 + 107 i

, so

x + i y = - 198 - 107 i

. Thus

x = - 198

and

y = - 107

, and

y - x = - 107 - (- 198) = 91

.

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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