Question

If$z = \cos \theta + i\sin \theta $, then the value of${z^n} + \dfrac{1}{{{z^n}}}$ will be
A.$\left( a \right){\text{ }}\sin 2n\theta $
B.$\left( b \right){\text{ }}2\sin n\theta $
C.$\left( c \right){\text{ }}2\cos n\theta $
D.$\left( d \right){\text{ }}\cos 2n\theta $

Answer 1

Hint: This type of problem is solved using a complex formula which is given by ${e^{i\theta }} = \cos \theta + i\sin \theta $. Replace the value of $z$ in the question by ${e^{i\theta }}$ and solve it. Also we should have knowledge of the inverse property which formula is given as \[\dfrac{1}{x} = {x^{ - 1}}\] .

Complete step-by-step answer:
The given equation in the problem is
${z^n} + \dfrac{1}{{{z^n}}} = ?$ … (1)
$z = \cos \theta + i\sin \theta $
Which further can be written using formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ as
$z = {e^{i\theta }}$
Replacing the value of $z$ in equation in \[\left( 1 \right)\]
${z^n} + \dfrac{1}{{{z^n}}} = {\left( {{e^{i\theta }}} \right)^n} + \dfrac{1}{{{{\left( {{e^{i\theta }}} \right)}^n}}}$
Using inverse formula \[\dfrac{1}{x} = {x^{ - 1}}\]it can further simplified as
$
   = {\left( {{e^{i\theta }}} \right)^n} + {\left( {{e^{i\theta }}} \right)^{ - n}} \\
   = {e^{in\theta }} + {e^{ - in\theta }} \\
 $
It can further solved using ${e^{i\theta }} = \cos \theta + i\sin \theta $ but here value of $\theta $ is $ n\theta $ then it can further written as ${e^{in\theta }} = \cos n\theta + i\sin n\theta $ then equation is
$
   = \left( {\cos n\theta + i\sin n\theta } \right) + \left( {\cos n\theta - i\sin n\theta } \right) \\
   = 2\cos n\theta \\
 $
So, the correct answer is “Option C”.

Note: we should remember the formula $z = {e^{i\theta }}$ and it can be manipulated using the value of $\theta $ which differs when power of $z$ is changed. These formulas are always manipulated in different types of formulas for different problems. Simplify these types of problems very carefully.