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JEE Main · 2004 · 24 Apr 2004 · Offline · Complex Numbers

If

z = x - i y

and

z^{\frac{1}{3}} = p + i q

, then

\frac{( \frac{x}{p} + \frac{y}{q} )}{( p ^{2} + q ^{2} )}

is equal to

(A)1
(B)-2✓ Correct
(C)2
(D)-1

Answer: (B)

Solution

Since

z^{1/3} = p + i q

, cube both sides to get

z = (p + i q)^{3} = p^{3} - 3 p q^{2} + i (3 p^{2} q - q^{3})

. Comparing with

z = x - i y

, the real part gives

x = p^{3} - 3 p q^{2} = p (p^{2} - 3 q^{2})

and the imaginary part gives

- y = 3 p^{2} q - q^{3}

, that is

y = q (q^{2} - 3 p^{2})

. Now

\frac{x}{p} + \frac{y}{q} = (p^{2} - 3 q^{2}) + (q^{2} - 3 p^{2}) = - 2 (p^{2} + q^{2})

, so dividing by

p^{2} + q^{2}

leaves

- 2

.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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