Question

How do you evaluate \[{e^{\dfrac{\pi }{2}i}} - {e^{\dfrac{{2\pi }}{3}i}}\] using trigonometric functions?

Answer 1

Hint: We are given an expression in complex exponential form and asked to evaluate the given expression using trigonometric functions. Here, you will need to use the Euler’s formula. Use this formula to change the exponential function into a trigonometric function.

Complete step by step solution:
Given the expression \[{e^{\dfrac{\pi }{2}i}} - {e^{\dfrac{{2\pi }}{3}i}}\]. We are asked to evaluate this expression.Let \[A = {e^{\dfrac{\pi }{2}i}} - {e^{\dfrac{{2\pi }}{3}i}}\] ………(i)
We will use here Euler’s formula. Euler’s formula is a formula which shows the relationship between complex exponential function and trigonometric functions. According to Euler’s formula,
\[{e^{i\theta }} = \cos \theta + i\sin \theta \] ……………(ii)
where cos and sin are trigonometric functions.

The first term of equation (i) is \[{e^{\dfrac{\pi }{2}i}}\].
Here \[\theta = \dfrac{\pi }{2}\]. Putting \[\theta = \dfrac{\pi }{2}\] in equation (ii) we get,
\[{e^{\dfrac{\pi }{2}i}} = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}\] ………………(iii)
The second term of equation (i) is \[{e^{\dfrac{{2\pi }}{3}i}}\].
Here \[\theta = \dfrac{{2\pi }}{3}\]. Putting \[\theta = \dfrac{{2\pi }}{3}\] in equation (ii) we get,
\[{e^{\dfrac{{2\pi }}{3}i}} = \cos \dfrac{{2\pi }}{3} + i\sin \dfrac{{2\pi }}{3}\]................(iv)

Putting the values from equation (iii) and (iv) in equation (i) we get,
\[A = \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right) - \left( {\cos \dfrac{{2\pi }}{3} + i\sin \dfrac{{2\pi }}{3}} \right)\]
\[ \Rightarrow A = 0 + i - \cos \dfrac{{2\pi }}{3} - i\sin \dfrac{{2\pi }}{3}\]
\[ \Rightarrow A = i - \cos \left( {\pi - \dfrac{\pi }{3}} \right) - i\sin \left( {\pi - \dfrac{\pi }{3}} \right)\]......................(v)
We know, \[\cos \left( {\pi - x} \right) = - \cos x\] and \[\sin \left( {\pi - x} \right) = \sin x\]. Using these formulas in equation (v) we get,
\[A = i - \left( { - \cos \dfrac{\pi }{3}} \right) - i\sin \dfrac{\pi }{3}\]
\[ \Rightarrow A = i + \cos \dfrac{\pi }{3} - i\sin \dfrac{\pi }{3}\]
\[ \Rightarrow A = i + \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i\]
\[ \therefore A = \dfrac{1}{2} + \left( {1 - \dfrac{{\sqrt 3 }}{2}} \right)i\]

Therefore, after evaluating we get \[{e^{\dfrac{\pi }{2}i}} - {e^{\dfrac{{2\pi }}{3}i}} = \dfrac{1}{2} + \left( {1 - \dfrac{{\sqrt 3 }}{2}} \right)i\].

Note: Remember that Euler’s formula is used for complex exponential functions and transform it in trigonometric functions. Also remember there are three main functions in trigonometry, these are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the main functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic formulas of trigonometry.