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Question: If \(\frac{1}{5 + 3\cos\theta}\), then the values of \((x + iy)^{1/3} = a + ib,\) and \(\frac{x}{a} ...

If 15+3cos⁡θ\frac{1}{5 + 3\cos\theta}, then the values of (x+iy)1/3=a+ib,(x + iy)^{1/3} = a + ib, and xa+yb\frac{x}{a} + \frac{y}{b}are.

A

4(a2+b2)4(a^{2} + b^{2})

B

4(a2−b2)4(a^{2} - b^{2})

C

4(b2−a2)4(b^{2} - a^{2})

D

{2i1+i}2=\left\{ \frac{2i}{1 + i} \right\}^{2} =

Answer

4(b2−a2)4(b^{2} - a^{2})

Explanation

Solution

μ2+λ2=4p2(p2−2)2+(5p2−1)2(4p2+1)2\mu^{2} + \lambda^{2} = \frac{4p^{2}(p^{2} - 2)^{2} + (5p^{2} - 1)^{2}}{(4p^{2} + 1)^{2}}⇒ =4p6+6p2+9p4+1(4p2+1)2=p4(4p2+1)+2p2(4p2+1)+(4p2+1)(4p2+1)2= \frac{4p^{6} + 6p^{2} + 9p^{4} + 1}{(4p^{2} + 1)^{2}} = \frac{p^{4}(4p^{2} + 1) + 2p^{2}(4p^{2} + 1) + (4p^{2} + 1)}{(4p^{2} + 1)^{2}}

Given series is G.P.

⇒=p4+2p2+14p2+1=(p2+1)24p2+1= \frac{p^{4} + 2p^{2} + 1}{4p^{2} + 1} = \frac{(p^{2} + 1)^{2}}{4p^{2} + 1} ⇒ z=3−4iz = 3 - 4i

⇒ z2=−7−24iz^{2} = - 7 - 24i

Equating real and imaginary parts, we get the required result.