Question

How do you multiply ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}$ in trigonometric form?

Answer 1

This question belongs to the topic polar system of the chapter complex numbers. For solving this question, we should know about Euler’s identity. The Euler’s identity is ${{e}^{ix}}=\cos x+i\sin x$. And, also for solving this question, we should know some formulas like the following:
${{i}^{2}}=-1$
${{e}^{a}}\times {{e}^{b}}={{e}^{a+b}}$.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to find the value of ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}$ in trigonometric form.
Let us first find the value of ${{e}^{\dfrac{\pi }{2}i}}$ and ${{e}^{3\dfrac{\pi }{2}i}}$ in trigonometric form separately.
To convert in trigonometric form, we will use the formula here is ${{e}^{ix}}=\cos x+i\sin x$
${{e}^{\dfrac{\pi }{2}i}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}$
As we know that the value of $\cos \dfrac{\pi }{2}$ is 0 and the value of $\sin \dfrac{\pi }{2}$ is 1. So, we can write the above equation as
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}=0+i\times 1=i$
Now, let us convert the second term.
${{e}^{3\dfrac{\pi }{2}i}}=\cos 3\dfrac{\pi }{2}+i\sin 3\dfrac{\pi }{2}$
As we know that $\cos \dfrac{3\pi }{2}=0$ and $\sin \dfrac{3\pi }{2}=-1$. So, after using these formulas in the above equation, we get
$\Rightarrow {{e}^{3\dfrac{\pi }{2}i}}=0+i\times (-1)=-i$
Now, we get that ${{e}^{\dfrac{\pi }{2}i}}=i$ and ${{e}^{3\dfrac{\pi }{2}i}}=-i$. Here, $i$ and $-i$ are the trigonometric forms of ${{e}^{\dfrac{\pi }{2}i}}$ and ${{e}^{3\dfrac{\pi }{2}i}}$ respectively.
So, we can find out the value of the term which is multiplication of ${{e}^{\dfrac{\pi }{2}i}}$ and ${{e}^{3\dfrac{\pi }{2}i}}$ that is ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}$.
Hence, ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}={{e}^{\dfrac{\pi }{2}i}}\times {{e}^{3\dfrac{\pi }{2}i}}$
As we have find the values of ${{e}^{\dfrac{\pi }{2}i}}$ and ${{e}^{3\dfrac{\pi }{2}i}}$, so we will put their values in the above equation, we get
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}={{e}^{\dfrac{\pi }{2}i}}\times {{e}^{3\dfrac{\pi }{2}i}}=i\times \left( -i \right)$
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}=-i\times i=-{{i}^{2}}$
As we know that ${{i}^{2}}=-1$, using this in the above equation, we get
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}=-{{i}^{2}}=-(-1)$
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}=1$

So, the trigonometric form of ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}$ is 1.

Note: At the time of solving this type of question, we should remember some value of trigonometric functions like sin, cos, etc. at angles $\dfrac{\pi }{2}$, $\pi$, $\dfrac{3\pi }{2}$, and $2\pi$.
$\cos \dfrac{\pi }{2}=0$; $\sin \dfrac{\pi }{2}=1$
$\cos \pi =-1$; $\sin \pi =0$
$\cos \dfrac{3\pi }{2}=0$; $\sin \dfrac{3\pi }{2}=-1$
$\cos 2\pi =1=\cos 0$; $\sin 2\pi =0=\sin 0$
Remember the above formulas to solve this type of question easily.
We can solve this question using an alternate method.
Using the formula ${{e}^{a}}\times {{e}^{b}}={{e}^{a+b}}$, we can write ${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}$ as
${{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}={{e}^{\dfrac{\pi }{2}i+3\dfrac{\pi }{2}i}}$
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}={{e}^{\dfrac{\left( 3+1 \right)\pi }{2}i}}={{e}^{\dfrac{4\pi }{2}i}}$
$\Rightarrow {{e}^{\dfrac{\pi }{2}i}}.{{e}^{3\dfrac{\pi }{2}i}}={{e}^{2\pi i}}$
Now, using the formula ${{e}^{ix}}=\cos x+i\sin x$ in the above equation, we can write
$\Rightarrow {{e}^{2\pi i}}=\cos 2\pi +i\sin 2\pi$
As we know that $\sin 2\pi =0$ and $\cos 2\pi =1$. So, the above equation can be written as
$\Rightarrow {{e}^{2\pi i}}=1+i\times 0=1$
Hence, we get the same value from this method too. So, one can use this method also.