Question

Express the following expression $\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}$ in the standard form.
A. $\left( {\dfrac{{1 - \cos \theta }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right) + i\left( {\dfrac{{ - 2\sin }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right)$
B. $\left( {\dfrac{{1 - \cos \theta }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right) + i\left( {\dfrac{{2\sin }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right)$
C. $\left( {\dfrac{{1 - \cos \theta }}{{2 + 2\cos \theta + 3{{\sin }^2}\theta }}} \right) + i\left( {\dfrac{{ - 2\sin }}{{2 + 2\cos \theta + 3{{\sin }^2}\theta }}} \right)$
D. $\left( {\dfrac{{1 + \cos \theta }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right) + i\left( {\dfrac{{ - 2\sin }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}} \right)$

Answer 1

Hint: In order to solve this question, first we will rationalize the given imaginary function in order to remove the imaginary term “i” from the denominator. Once an imaginary term is removed from the denominator, we will then use the algebraic identity in order to simplify the term and then separate the real and imaginary part of the term to bring it in standard form.

Complete step-by-step answer:
Given term is $\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}$
We know that if a complex number is given as $\dfrac{1}{{a + ib}}$ then for rationalizing we have to multiply it by $\dfrac{{a - ib}}{{a - ib}}$
By rationalizing the given term, we have
$
   = \dfrac{1}{{\left( {1 - \cos \theta } \right) + \left( {2i\sin \theta } \right)}} \times \dfrac{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}}{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}} \\
  \left[ {\because \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \\
   = \dfrac{{\left( {1 - \cos \theta } \right) - \left( {i2\sin \theta } \right)}}{{{{\left( {1 - \cos \theta } \right)}^2} - {{\left( {2i\sin \theta } \right)}^2}}} \\
   = \dfrac{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}}{{1 - 2\cos \theta + {{\cos }^2}\theta - ( - 1)4{{\sin }^2}\theta }}{\text{ }}\left[ {\because {i^2} = - 1} \right] \\
   = \dfrac{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}}{{1 - 2\cos \theta + {{\cos }^2}\theta + 4{{\sin }^2}\theta }}{\text{ }} \\
   = \dfrac{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}}{{1 - 2\cos \theta + 1 + 3{{\sin }^2}\theta }}{\text{ }}\left[ {\because {{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right] \\
   = \dfrac{{\left( {1 - \cos \theta } \right) - i\left( {2\sin \theta } \right)}}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }} \\
 $
Now we write it in standard form which is given as $x + iy$
$ \Rightarrow \dfrac{{\left( {1 - \cos \theta } \right)}}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}{\text{ }} + i\dfrac{{ - i\left( {2\sin \theta } \right)}}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}$

So, the correct answer is “Option C”.


Note: In order to solve these types of questions, remember the basic algebraic identities and some complex number properties such as conjugate of complex numbers. Also remember how to rationalize a fraction number. Also remember other properties of complex numbers which will help in further problems such as the value of ${i^4} = 1$ and complex numbers can be represented in exponential form or in sine-cosine form.