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Question: How do you find the nth root of -1?...

How do you find the nth root of -1?

Explanation

Solution

Hint : In the given question, we have been asked to find out the nth root of -1 i.e. nth root of the complex number. In order to find the nth root, first we will use the concept of nth root of a complex number and then we will use De Moivre’s theorem which states that If r(cos⁡θ+isin⁡θ)r\left( \cos \theta +i\sin \theta \right) is the given complex number in the polar form, then (r(cos⁡θ+isin⁡θ))n=rn(cos⁡nθ+isin⁡nθ){{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right) . Later we will start writing the equation of a complex number in a polar form and then according to the values we will substitute the angles and solve step by step to get the nth root of -1.

Complete step-by-step answer :
The complex number ‘z’ in polar form is given by,
z=r(cos⁡θ+isin⁡θ)z=r\left( \cos \theta +i\sin \theta \right) , then
Raising the power ‘n’ to both the sides, we will obtain
zn=(r(cos⁡θ+isin⁡θ))n{{z}^{n}}={{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}
Using the de moivre’s theorem which states that,
If r(cos⁡θ+isin⁡θ)r\left( \cos \theta +i\sin \theta \right) is the given complex number in the polar form, then
(r(cos⁡θ+isin⁡θ))n=rn(cos⁡nθ+isin⁡nθ){{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)
Let zn{{z}^{n}} = -1,
Therefore, substituting the value, we will get
−1=1(cos⁡nθ+isin⁡nθ)-1=1\left( \cos n\theta +i\sin n\theta \right)
Substituting cos⁡nθ=cos⁡π\cos n\theta =\cos \pi and sin⁡nθ=sin⁡π\sin n\theta =\sin \pi , we will get
−1=1(cos⁡π+isin⁡π)-1=1\left( \cos \pi +i\sin \pi \right)
Using the trigonometric ratios table, we know the values of,
cos⁡π=−1\cos \pi =-1 , and sin⁡π=0\sin \pi =0
Substituting the values, we will get
−1=1(−1+0)=−1-1=1\left( -1+0 \right)=-1 .
Therefore,
‘z’ the nth root of -1 will be given by,
nthroot of −1=cos⁡(πn)+isin⁡(πn){{n}^{th}}root\ of\ -1=\cos \left( \dfrac{\pi }{n} \right)+i\sin \left( \dfrac{\pi }{n} \right)
Hence, this will be the required answer to the question.

Note : As we know that the complex numbers are the given points in the complex plane and the nth roots always have the magnitude of 1. Therefore, all the nth roots lie inside the circle of radius of 1 in the given complex plane. In order to solve these types of problems or the questions, students must need to remember the theorem of De Moivre’s.