Question

The real value of $\theta$ for which the expression $\left( {1 + i\cos \theta } \right){\left( {1 - 2i\cos \theta } \right)^{ - 1}}$ is purely imaginary is
$1)$ $n\pi$
$2)$ $n\pi \pm \dfrac{\pi }{6}$
$3)$ $n\pi \pm \dfrac{{2\pi }}{3}$
$4)$ $n\pi \pm \dfrac{\pi }{4}$

Answer 1

We shall discuss the complex numbers. While squaring a number results in a negative result, it can be known as imaginary numbers and it is denoted by a letter $i$. A complex number is a number that is formed due to the combination of a real number and an imaginary number.
Examples for complex numbers are $2 + 2i$, $4 - 5i$ .
In standard form, the complex number is written as,
$z = a + bi$
Where $a = \operatorname{Re} Z$(i.e. $a$ is the real part of $Z$)
$b = \operatorname{Im} Z$(i.e. $b$ is the imaginary part of $Z$)
It is given that the above complex number is purely imaginary.
That is, if a complex number has no real part, then it is said to be purely imaginary.
Formula used:
$\left( {a + bi} \right)\left( {a - bi} \right) = {a^2} + {b^2}$

Complete step-by-step solution:
The given expression is $\left( {1 + i\cos \theta } \right){\left( {1 - 2i\cos \theta } \right)^{ - 1}}$
Let,
$z = \left( {1 + i\cos \theta } \right){\left( {1 - 2i\cos \theta } \right)^{ - 1}}$
It can be written as,
$z = \dfrac{{1 + icos\theta }}{{1 - 2icos\theta }}$
Taking conjugate, we get
$z = \dfrac{{(1 + icos\theta )(1 + 2icos\theta )}}{{(1 - 2icos\theta )(1 + 2icos\theta )}}$
Now, applying the formula on the denominator, we have
$z = \dfrac{{1 - 2co{s^2}\theta + i3cos\theta }}{{1 + 4co{s^2}\theta }}$
$z = \dfrac{{1 - 2co{s^2}\theta }}{{1 + 4co{s^2}\theta }} + i(\dfrac{{3cos\theta }}{{1 + 4co{s^2}\theta }})$
It is given that the above complex number is purely imaginary.
That is, if a complex number has no real part, then it is said to be purely imaginary.
That is,
$\operatorname{Re} (z) = 0$
$\dfrac{{1 - 2co{s^2}\theta }}{{1 + 4co{s^2}\theta }} = 0$
$1 - 2co{s^2}\theta = 0$
$\;1 = 2co{s^2}\theta$
$\;co{s^2}\theta = \dfrac{1}{2}$
Taking square roots on both sides,
$\cos \theta = \dfrac{1}{{\sqrt 2 }}$
$\theta = {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$
$\theta = n\pi \pm \dfrac{\pi }{4}$ , where $n \in N$ .

Note: A complex number is a number that is formed due to the combination of a real number and an imaginary number. Examples of complex numbers are $2 + 2i$, $4 - 5i$ .
Here, in our question, it is given that the above complex number is purely imaginary.
That is, if a complex number has no real part, then it is said to be purely imaginary.
If a complex number has no imaginary part, then it is said to be purely real.