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Question

Question: How is \({{e}^{i\pi }}=-1\)? And what is \({{e}^{i\dfrac{\pi }{4}}}\)?...

How is eiπ=1{{e}^{i\pi }}=-1? And what is eiπ4{{e}^{i\dfrac{\pi }{4}}}?

Explanation

Solution

We can solve the above given question by applying the Euler’s formula. Euler’s formula is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function.

Complete step by step answer:
Euler’s formula states that for any real number xx,eix=cosx+isinx{{e}^{ix}}=\cos x+i\sin x
Where ee is the base of the natural logarithm, ii is the imaginary unit, and cos\cos and sin\sin are the trigonometric functions sine and cosine respectively.
This complex exponential function is sometimes denoted by cis xcis\text{ x}. The formula is still valid if  x\text{ x} is a complex number.
Now according to the given question we have to prove that, eiπ=1{{e}^{i\pi }}=-1
Now from the Euler’s formula we can solve it.
As we have been already discussed earlier Euler’s formula is, eix=cosx+isinx{{e}^{ix}}=\cos x+i\sin x
eiπ=cosπ+isinπ\Rightarrow {{e}^{i\pi }}=\cos \pi +i\sin \pi
Now by using the trigonometric values of sinπ=0 and cosπ=-1\sin \pi =0\text{ and cos}\pi \text{=-1}, we get,
eiπ=1+0\Rightarrow {{e}^{i\pi }}=-1+0
eiπ=1\Rightarrow {{e}^{i\pi }}=-1
Hence, proved
Now, as it have been asked in the question now we have to calculate eiπ4{{e}^{i\dfrac{\pi }{4}}}
Now we can solve this also by using the Euler’s formula.
As we have been already discussed earlier Euler’s formula is eix=cosx+isinx{{e}^{ix}}=\cos x+i\sin x
Now by using the Euler’s formula we can solve this one.
eiπ4=cosπ4+isinπ4{{e}^{i\dfrac{\pi }{4}}}=\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}
Now by using the trigonometric values of cosπ4=12 and sinπ4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\text{ and sin}\dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} we can solve the complex equation. Now substitute the values of sine and cosine in the complex equation. By substituting the values we get,
eiπ4=12 +i12 {{e}^{i\dfrac{\pi }{4}}}=\dfrac{1}{\sqrt{2}}\text{ }+i\dfrac{1}{\sqrt{2}}\text{ }
eiπ4=1+i2 \Rightarrow {{e}^{i\dfrac{\pi }{4}}}=\dfrac{1+i}{\sqrt{2}}\text{ }

Therefore we can conclude that eiπ4=1+i2 {{e}^{i\dfrac{\pi }{4}}}=\dfrac{1+i}{\sqrt{2}}\text{ } .

Note: We should be careful while doing the complex numbers. We should be well aware of the complex numbers and Euler’s formula and its usage. The Euler’s formula states that the value of eiθ{{e}^{i\theta }} is cosθ+isinθ\cos \theta +i\sin \theta where θ\theta is the argument we can say that any complex number can be expressed as z=a+ib=reiθz=a+ib=r{{e}^{i\theta }} where r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}} .