Question

How do you evaluate ${{e}^{\dfrac{\pi }{4}i}}-{{e}^{\dfrac{\pi }{3}i}}$ using trigonometric functions$?$

Answer 1

we know that ${{e}^{\theta i}}$ can be written in trigonometric functions as $\cos \left( \theta \right)+i\sin \left( \theta \right)$ similarly here we have to change the value of $\theta$ to $\dfrac{\pi }{4}$ and $\dfrac{\pi }{3}$. To get the values of ${{e}^{\dfrac{\pi }{4}i}}$ and ${{e}^{\dfrac{\pi }{3}i}}$ to get the values as $\cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right)$ and $\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right)$ then find the values of trigonometric values of $\cos \left( \dfrac{\pi }{4} \right)$ , $\sin \left( \dfrac{\pi }{4} \right)$,$\cos \left( \dfrac{\pi }{3} \right)$ and $\sin \left( \dfrac{\pi }{3} \right)$. Now substitute the values in the problem to get the simplified form.

Complete step by step solution:
We can write ${{e}^{\dfrac{\pi }{4}i}}$ as follows,
${{e}^{\dfrac{\pi }{4}i}}$
$\Rightarrow \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right)$
We know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$
$\Rightarrow \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}$
$\Rightarrow \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}$
Similarly,
${{e}^{\dfrac{\pi }{3}i}}$
$\Rightarrow \cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right)$
We know that $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$
$\Rightarrow \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$

Hence we can substitute the values that we have obtained from the above calculations to the given question as below,
${{e}^{\dfrac{\pi }{4}i}}-{{e}^{\dfrac{\pi }{3}i}}$
$\Rightarrow \left( \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2} \right)-\left( \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right)$
$\Rightarrow \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$
$\Rightarrow \dfrac{\sqrt{2}}{2}-\dfrac{1}{2}+i\left( \dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2} \right)$
$\Rightarrow \dfrac{\sqrt{2}-1}{2}+i\left( \dfrac{\sqrt{2}-\sqrt{3}}{2} \right)$
From this the problem provided, it is ${{e}^{\dfrac{\pi }{4}i}}-{{e}^{\dfrac{\pi }{3}i}}$ can be written as $\dfrac{\sqrt{2}-1}{2}+i\left( \dfrac{\sqrt{2}-\sqrt{3}}{2} \right)$ using the trigonometric functions.

Note: In this type of problems we have to be well known with the conversion of ${{e}^{\theta i}}$ to $\cos \left( \theta \right)+i\sin \left( \theta \right)$and the main part in this question is to convert the trigonometric values of basic sine and cosine values. Here in conversion of the form ${{e}^{\theta i}}$ we will be having a real part and an imaginary part. So it is very important to take the real terms together and perform the arithmetic operations and take the imaginary terms together and perform the arithmetic operations on them to get a simplified form of complex number. Here we should always remember to represent the obtained complex number in the standard form of $a+ib$ so that we mention real part and imaginary part separately.