Question

Find the real values of $\theta $ for which $\left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)$ is purely real.

Answer 1

Hint: At first, we rationalise the denominator by $\dfrac{\left( 1+i\cos \theta \right)\left( 1+2i\cos \theta \right)}{\left( 1-2i\cos \theta \right)\left( 1+2i\cos \theta \right)}$ . Then, we simplify the expression to $\dfrac{1-2{{\cos }^{2}}\theta +3i\cos \theta }{1+4{{\cos }^{2}}\theta }$ . After that, we separate the real and imaginary parts as $\dfrac{1-2{{\cos }^{2}}\theta }{1+4{{\cos }^{2}}\theta }+\dfrac{3i\cos \theta }{1+4{{\cos }^{2}}\theta }$ . Then, we equate the imaginary part to zero and get $\cos \theta =0$ which gives the real values of $\theta $ .

Complete step by step solution:
The expression that we are given in this problem is,
$\left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)$
We need to simplify this expression in order to arrive at a conclusion. We can simplify it by rationalising it. We rationalise a fraction, say $\dfrac{x+y}{a-b}$ by multiplying $a+b$ in the numerator and the denominator as in $\dfrac{\left( x+y \right)\left( a+b \right)}{\left( a-b \right)\left( a+b \right)}$ . Then, doing so with the given fraction, we get,
$\Rightarrow \left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)=\dfrac{\left( 1+i\cos \theta \right)\left( 1+2i\cos \theta \right)}{\left( 1-2i\cos \theta \right)\left( 1+2i\cos \theta \right)}$
Simplifying the multiplication, we get,
$\Rightarrow \left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)=\dfrac{1+2i\cos \theta +i\cos \theta +\left( 2i\cos \theta \times i\cos \theta \right)}{1-{{\left( 2i\cos \theta \right)}^{2}}}$
We know that ${{i}^{2}}=-1$ . This gives,
$\Rightarrow \left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)=\dfrac{1+3i\cos \theta +2\left( -1 \right){{\cos }^{2}}\theta }{1-4\left( -1 \right){{\cos }^{2}}\theta }$
Further simplifying, we get,
$\Rightarrow \left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)=\dfrac{1-2{{\cos }^{2}}\theta +3i\cos \theta }{1+4{{\cos }^{2}}\theta }$
Separating the real and the imaginary parts, we get,
$\Rightarrow \left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)=\dfrac{1-2{{\cos }^{2}}\theta }{1+4{{\cos }^{2}}\theta }+\dfrac{3i\cos \theta }{1+4{{\cos }^{2}}\theta }$
Now, any expression with an $i$ multiplied to it is an imaginary expression. In the above operation, the term $3\cos \theta $ had $i$ multiplied with it, so we separated it. So, in order to make the expression purely real, the imaginary part should vanish, or in other words should be zero. This means that, if we equate the expression multiplied with $i$ to zero, we get the condition for the entire expression to be purely real. So,
$\begin{align}
  & \Rightarrow \dfrac{3\cos \theta }{1+4{{\cos }^{2}}\theta }=0 \\
 & \Rightarrow 3\cos \theta =0 \\
 & \Rightarrow \cos \theta =0 \\
\end{align}$
Now, $\cos \theta $ is zero for $\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2},...$ . So, for $\theta =\dfrac{n\pi }{2}$ where, $n=1,3,5,...$ $\cos \theta $ is zero.
Thus, we can conclude that the real values of $\theta $ for which $\left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)$ is purely real are $\theta =\dfrac{n\pi }{2}$ where, $n=1,3,5,...$.

Note: We should remember to write the general expression for $\theta $ and not just simply \[\dfrac{\pi }{2}\] . Also, we can solve this problem in another way. But this special method will be applicable only for this problem. If we can eliminate all the imaginary terms in the expression $\left( \dfrac{1+i\cos \theta }{1-2i\cos \theta } \right)$ , we get the answer. This can be done by equating $\cos \theta $ to zero, which is what we have done.