SATHEE: Complex Numbers 2 Question 33

Complex Numbers 2 Question 33

34. If the expression $\frac{\sin \frac{x}{2} + \cos \frac{x}{2} - i \tan (x)}{1 + 2 i \sin \frac{x}{2}}$

is real, then the set of all possible values of $x$ is… .

$(1987, 2 M)$

True/False

Show Answer

Answer:

Correct Answer: 34. $(x = 3$ and $y = - 1)$

$46. A + i B = \frac{1}{21 + 3 \cos^{2} \frac{θ}{2}} - i \frac{\cot (θ / 2)}{1 + 3 \cos^{2} (θ / 2)}$

Solution:

$\frac{\sin \frac{x}{2} + \cos \frac{x}{2} - i \tan x}{1 + 2 i \sin \frac{x}{2}} \in R$

$= \frac{\sin \frac{x}{2} + \cos \frac{x}{2} - i \tan x 1 - 2 i \sin \frac{x}{2}}{1 + 4 \sin^{2} \frac{x}{2}}$

Since, it is real, so imaginary part will be zero.

$\begin{aligned} ∴ - 2 \sin \frac{x}{2} \sin \frac{x}{2} + \cos \frac{x}{2} - \tan x = 0 \\ \Rightarrow 2 \sin \frac{x}{2} \sin \frac{x}{2} + \cos \frac{x}{2} \cos x + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 0 \\ \Rightarrow \sin \frac{x}{2} \sin \frac{x}{2} + \cos \frac{x}{2} \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} + \cos \frac{x}{2} = 0 \\ ∴ \sin \frac{x}{2} = 0 \\ \Rightarrow \sin \frac{x}{2} + \cos \frac{x}{2} \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} + \cos \frac{x}{2} = 0 \end{aligned}$

On dividing by $\cos^{3} \frac{x}{2}$ , we get

$\begin{aligned} \tan \frac{x}{2} + 1 1 - \tan^{2} \frac{x}{2} + 1 + \tan^{2} \frac{x}{2} = 0 \\ \Rightarrow \tan^{3} \frac{x}{2} - \tan \frac{x}{2} - 2 = 0 \\ Let \tan \frac{x}{2} = t \\ and f (t) = t^{3} - t - 2 \\ Then, f (1) = - 2 < 0 \\ and f (2) = 4 > 0 \end{aligned}$

Thus, $f (t)$ changes sign from negative to positive in the interval $(1, 2)$ .

$∴$ Let $t = k$ be the root for which

$\begin{array}{lc} f (k) = 0 and k \in (1, 2) \\ ∴ & t = k or \tan \frac{x}{2} = k = \tan α \\ \Rightarrow & x / 2 = n π + α \\ \Rightarrow & x = 2 n π + 2 α, α = \tan^{- 1} k, where k \in (1, 2) \\ or x = 2 n π \end{array}$