
How do you find the nth root of -1?
Answer
527.1k+ views
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 \[r\left( \cos \theta +i\sin \theta \right)\] is the given complex number in the polar form, then \[{{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)\] . 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,
\[z=r\left( \cos \theta +i\sin \theta \right)\] , then
Raising the power ‘n’ to both the sides, we will obtain
\[{{z}^{n}}={{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}\]
Using the de moivre’s theorem which states that,
If \[r\left( \cos \theta +i\sin \theta \right)\] is the given complex number in the polar form, then
\[{{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)\]
Let \[{{z}^{n}}\] = -1,
Therefore, substituting the value, we will get
\[-1=1\left( \cos n\theta +i\sin n\theta \right)\]
Substituting \[\cos n\theta =\cos \pi \] and \[\sin n\theta =\sin \pi \] , we will get
\[-1=1\left( \cos \pi +i\sin \pi \right)\]
Using the trigonometric ratios table, we know the values of,
\[\cos \pi =-1\] , and \[\sin \pi =0\]
Substituting the values, we will get
\[-1=1\left( -1+0 \right)=-1\] .
Therefore,
‘z’ the nth root of -1 will be given by,
\[{{n}^{th}}root\ of\ -1=\cos \left( \dfrac{\pi }{n} \right)+i\sin \left( \dfrac{\pi }{n} \right)\]
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.
Complete step-by-step answer:
The complex number ‘z’ in polar form is given by,
\[z=r\left( \cos \theta +i\sin \theta \right)\] , then
Raising the power ‘n’ to both the sides, we will obtain
\[{{z}^{n}}={{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}\]
Using the de moivre’s theorem which states that,
If \[r\left( \cos \theta +i\sin \theta \right)\] is the given complex number in the polar form, then
\[{{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)\]
Let \[{{z}^{n}}\] = -1,
Therefore, substituting the value, we will get
\[-1=1\left( \cos n\theta +i\sin n\theta \right)\]
Substituting \[\cos n\theta =\cos \pi \] and \[\sin n\theta =\sin \pi \] , we will get
\[-1=1\left( \cos \pi +i\sin \pi \right)\]
Using the trigonometric ratios table, we know the values of,
\[\cos \pi =-1\] , and \[\sin \pi =0\]
Substituting the values, we will get
\[-1=1\left( -1+0 \right)=-1\] .
Therefore,
‘z’ the nth root of -1 will be given by,
\[{{n}^{th}}root\ of\ -1=\cos \left( \dfrac{\pi }{n} \right)+i\sin \left( \dfrac{\pi }{n} \right)\]
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

