Question

Express \[\dfrac{1}{{1 + \cos \theta - i\sin \theta }}\]in the form of \[a + ib\]

Answer 1

Hint: Firstly rationalize the denominator. Using various algebraic identities related to squares simplify the denominator. Use the trigonometric identity that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]. Hence separate the real and imaginary parts and you get the answer.

Complete step-by-step answer:
To rationalize the denominator means to eliminate any radical expressions in the denominator such as square roots and cube roots. The key idea is to multiply the original dfraction by an appropriate value, such that after simplification, the denominator no longer contains radicals.
A complex number is a number that can be expressed in the form \[x + iy\] where \[x\] and \[y\] are real numbers and \[i\] s a symbol called the imaginary unit, and satisfying the equation \[{i^2} = - 1\] . Because no "real" number satisfies this equation \[i\] was called an imaginary number. For a complex number \[x + iy\] , \[x\] is called the real part and \[y\] is called the imaginary part.
We have \[\dfrac{1}{{\left( {1 + \cos \theta } \right) - i\sin \theta }}\]
On rationalizing the denominator we get
\[\dfrac{1}{{\left( {1 + \cos \theta } \right) - i\sin \theta }} \times \dfrac{{\left( {1 + \cos \theta } \right) + i\sin \theta }}{{\left( {1 + \cos \theta } \right) + i\sin \theta }}\]
Using the identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] in the denominator we get
\[ = \dfrac{{\left( {1 + \cos \theta } \right) + i\sin \theta }}{{{{\left( {1 + \cos \theta } \right)}^2} - {i^2}{{\sin }^2}\theta }}\]
Opening the square term in the denominator using the identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] and also putting value of \[{i^2} = 1\] we get
\[ = \dfrac{{\left( {1 + \cos \theta } \right) + i\sin \theta }}{{1 + 2\cos \theta + {{\cos }^2}\theta + {{\sin }^2}\theta }}\]
Since we know that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] we get
\[ = \dfrac{{1 + \cos \theta + i\sin \theta }}{{2 + 2\cos \theta }}\]
Now taking \[2\] common from the denominator and separating the real and imaginary parts we get
\[ = \dfrac{{1 + \cos \theta }}{{2\left( {1 + \cos \theta } \right)}} + i\dfrac{{\sin \theta }}{{2\left( {1 + \cos \theta } \right)}}\]
Cancelling the same terms in the real part we get
\[ = \dfrac{1}{2} + i\dfrac{{\sin \theta }}{{2\left( {1 + \cos \theta } \right)}}\] which is in the form of \[a + ib\].

Note: Firstly rationalize the denominator. Using various algebraic identities related to squares simplify the denominator. Wisely choose the identity to be used in order to simplify the denominator , mostly \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] is used. Keep in mind to separate the real and imaginary part at the end to represent it in the form of a complex number.