Question

If $x+iy=\dfrac{1}{(1+cosθ)+isinθ}$,then the value of $x^2+1$ is ?

• A. $\dfrac{7}{4}$
• B. $\dfrac{13}{4}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{9}{4}$
• E. $\dfrac{5}{4}$

Answer 1

Answer: (E)

Given: $x + iy = \frac{1}{(1+\cos\theta) + i\sin\theta}$

First, rationalize the denominator: $$x + iy = \frac{(1+\cos\theta) - i\sin\theta}{(1+\cos\theta)^2 + \sin^2\theta}$$

Simplify the denominator: $$(1+\cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + sin^2\theta = 2 + 2\cos\theta$$

Thus: $$x + iy = \frac{1+\cos\theta}{2(1+\cos\theta)} - i\frac{\sin\theta}{2(1+\cos\theta)} = \frac{1}{2} - i\frac{\sin\theta}{2(1+\cos\theta)}$$

Therefore: $$x = \frac{1}{2}$$ $$y = -\frac{\sin\theta}{2(1+\cos\theta)}$$

Now compute $x^2 + 1$: $$x^2 + 1 = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}$$

The correct answer is (E) $\frac{5}{4}$.

Answer 2

Given that:

 $x+iy=1/(1+cosθ)+isinθ$

$=\dfrac{1}{1+2cos^2(θ/2)+i.2sin(θ/2).cos(θ/2)}$

$=\dfrac{1}{1+2cos^2(θ/2)}.\dfrac{1}{e^{i.(θ/2)}}$

$=\dfrac{1}{1+2cos^2(θ/2)}.e^{-i.(θ/2)}$

$=\dfrac{1}{1+2cos^2(θ/2)}.(cos(θ/2)-isin(θ/2))$

$=\dfrac{1}{2}-itan(θ/2)$

Hence, now we can write that  $x=\dfrac{1}{2}$

So, $x^2+1=(\dfrac{1}{2})^2+1=\dfrac{5}{4}$ (_Ans.)