Question

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

Answer 1

Hint: 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.