Question

The locus of the point whose position vector is given by (a + ib)⁵ + (b + ia)⁵ (where a, b are real parameters) is-

• A. y = x
• B. y = – x
• C. y = mx, m Ī R
• D. Not defined

Answer 1

Answer: (A)

y = x

Answer 2

Sol. Let ; a = r cos q, b = r sin q

Then (a + ib)⁵ + (b + ia)⁵

= r⁵ {(cos q + i sin q)⁵ + (sin q + i cos q)⁵

= r⁵ $\left\lbrack (\cos 5\theta + i\sin 5\theta) + \left\{ \cos\left( \frac{\pi}{2} - \theta \right) + i\sin\left( \frac{\pi}{2} - \theta \right) \right\}^{5} \right\rbrack$=r⁵

$\left\lbrack (\cos 5\theta + i\sin 5\theta) + \cos 5\left( \frac{\pi}{2} - \theta \right) + i\sin 5\left( \frac{\pi}{2} - \theta \right) \right\rbrack$ = r⁵

[(cos5q + i sin 5q) (1 + i)].

This is a complex number whose real and imaginary parts are equal. So locus of such a point will be y = x.