 Hello everyone, Myself A.S. Falmari, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapur. In this video, we are going to learn how to find out the roots of a complex number using de Moivre's theorem. Learning outcome, at the end of the session, students will be able to find the roots of a complex number using de Moivre's theorem. Let us start this lecture with a theorem. Let w is equal to r into the bracket cos theta plus i sin theta be any non-zero complex number. Now actually w is a complex number which is given in a polar form. Then the n nth roots of a complex number w are given by z is equal to the nth root of a complex number w is denoted by the notation w to the power one by n and is given by the formula r to the power one by n into the bracket cos of 2k pi plus theta upon n plus i into sin of 2k pi plus theta upon n where k is equal to 0 comma 1 comma 2 so on up to n minus 1. Now working rule, now the step number one, the complex number was nth root we have to find out if that complex number is given in polar form then we directly go for step two. If that complex number is given in Cartesian form then we have to perform this step number one. This step number one is nothing but writing the given complex number in terms of a polar form. Suppose we have given a complex number in terms of Cartesian form like x plus i y and we know that its polar form is given by x plus i y is equal to r into the bracket cos theta plus i sin theta where two quantities we have to determine r and theta and that are given by r is equal to under root of x square plus y square and this theta is equal to tan inverse of y by x and once we express the given complex number in terms of polar form then we directly apply the theorem then by the above theorem the nth root of a complex number is given by z that is equal to now the nth root of the given complex number x plus i y is denoted by x plus i y to the power 1 by n and it is given as r to the power 1 by n inside the bracket cos of 2k pi plus theta upon n plus i sin of 2k pi plus theta upon n where k is equal to 0 1 2 3 comma so on up to n minus 1. Pause this video and write down the complex numbers 1 minus 1 i minus i in polar form. I hope all of you have written the answer now we know that these are the standard complex number which has the standard polar form now one has the polar form cos 0 plus i sin 0 minus one has the polar form cos pi plus i sin pi i has the polar form cos of pi by 2 plus i sin pi by 2 and minus i has the polar form cos of pi by 2 minus i into sin of pi by 2 consider the first example find the cube roots of unity solution now here the provided complex number is unity unity means one just we have discussed a complex number one can be written in polar form as cos 0 plus i sin 0 now here we have to find out the cube root of one pi the above theorem cube root of this unity is given by z is equal to now the cube root of unity is denoted by 1 to the power 1 by 3 and it is given as cos of 2k pi plus now original angle 0 divided by here the value of n is 3 divided by 3 plus again i into sin of 2k pi plus 0 upon 3 and now we know that 2k pi plus 0 is 2k pi therefore we get z as cos of 2k pi by 3 plus i sin of 2k pi by 3 where k equal to 0 comma 1 comma 2 let us find out the values of z for various values of k for k is equal to 0 when we put k is equal to 0 here we get a complex number and we denote it as z 0 is equal to cos of 0 plus i sin 0 we know that cos 0 has the value 1 and sin 0 is 0 therefore z 0 is becomes 1 now next value of k is 1 now putting k is equal to 1 in this formula we get it as and we denote this complex number by z 1 is equal to cos of 2 pi by 3 plus i sin of 2 pi by 3 is equal to 2 pi by 3 is the actually standard angle now we can write it as cos of pi minus pi by 3 plus i sin of pi minus pi by 3 now we know the results that cos of pi minus theta is minus cos theta and sin of pi minus theta is simply sin theta using these two results in this equation we get it as minus cos of pi by 3 plus i sin of pi by 3 that is equal to and we can write this minus as it is we know that cos of pi by 3 is 1 by 2 plus i as it is and sin of pi by 3 is root 3 by 2 now this is the next required value of cube root of unity and last one for k equal to 2 we will denote the complex number by z 2 now putting k equal to 2 here we get it as 2 into 2 4 now we get z 2 is equal to cos of 4 pi by 3 plus i sin of 4 pi by 3 that is equal to now we can write this angle 4 pi by 3 as cos of pi plus pi by 3 plus i into sin of pi plus pi by 3 again we know that the result cos of pi plus theta equal to minus cos theta and sin of pi plus theta is minus sin theta now using these two results in this equation we get cos of pi plus pi by 3 as minus cos of pi by 3 and plus i as it is and sin of pi plus pi by 3 as minus of sin pi by 3 and this i as it is and writing minus as it is cos of pi by 3 is nothing but 1 by 2 minus i as it is and sin of pi by 3 is root 3 by 2 now these are the z naught z 1 z 2 are the required cube roots of unity let us consider one more example if omega is a cube root of unity then prove that 1 plus omega plus omega square is equal to 0 and also prove that 1 minus omega bracket raised to 6 is nothing but minus 27 solution just we have calculated the cube roots of unity i am not going to calculate again the same roots we know that the three roots are z naught z 1 z 2 where z naught is 1 and z 1 is minus 1 by 2 plus i into root 3 by 2 we will denote it as omega and z 2 is equal to minus 1 by 2 minus i root 3 by 2 we will denote it as omega square now adding these three numbers we get 1 plus omega plus omega square is equal to substituting these two values we get 1 plus inside the bracket minus 1 by 2 plus i into root 3 by 2 plus inside the bracket minus 1 by 2 minus i into root 3 by 2 now here we can see that the this term is plus i root 3 by 2 and this one is minus i into root 3 by 2 they are the same terms with opposite sign we can remove them now we can write this one as it is now here it is minus 1 by 2 here is also minus 1 by 2 minus 1 by 2 minus 1 by 2 is minus 1 and 1 minus 1 is nothing but 0 now the next thing we have to calculate or we have to show that is 1 minus omega whole bracket raise to 6 is nothing but minus 27 now the consider 1 minus omega bracket raise to 6 the is equal to now i can write here this power 6 as 1 minus omega bracket raise to 2 and whole bracket raise to 3 we have split it that 6 as power 2 and power 3 now expanding internal bracket we get it as 1 minus 2 omega plus omega square by using the property a minus b bracket square is a square minus 2 ab plus b square now that is equal to now just we have seen that the value of 1 plus omega plus omega square is 0 so the value of 1 plus omega square is nothing but minus omega substituting the value of 1 plus omega square as minus omega here and writing this minus 2 omega as it is and the bracket raise to 3 minus omega minus 2 omega is minus 3 omega and whole bracket raise to 3 and taking the cube of this one minus 3 cube is minus 27 and omega cube is here omega cube hence it is proved in general the nth roots of unity has three standard properties the first one is the sum of all the n nth roots of unity is always 0 that is if 1 omega omega square so on up to omega to the power n minus 1 are the n nth roots of unity then 1 plus omega plus omega square so on up to plus omega to the power n minus 1 is always 0 one more important property is that the product of all the n nth roots of unity is always minus 1 to the power n minus 1 that is if we choose 1 omega omega square so on up to omega raise to n minus 1 are the n nth roots of unity then the value of the product 1 into omega into omega square into so on up to into omega to the power n minus 1 is nothing but minus 1 to the power n minus 1 and the final last property is that all the n nth roots of unity are always in geometric progression