 everyone, myself AS Falmari, Assistant Professor, Department of Humanities and Sciences, Valchand Institute of Technology, Sholapur. In the last two videos we have discussed how to find out the roots of a complex number. In this video also, we will discuss how to find out the roots of an equation using DeMoevers theorem. Learning outcome, at the end of this session, students will be able to find roots of an equation using DeMoevers theorem. Consider the first example. Use DeMoevers theorem to solve x to the power 4 minus x cube plus x square minus x plus 1 equal to 0. Solution. In order to obtain the roots of any algebraic equation using DeMoevers theorem, the given equation must be converted into the form of x to the power n plus minus alpha equal to 0, where alpha is any real or complex number. To convert this given polynomial in the form of x to the power n plus minus alpha, we have to multiply it by a suitable factor. Now here, we multiply by a suitable factor x plus 1 to the given polynomial, we get this equation. Now multiplying by x to all the terms of the second bracket, we get here x into x to the power 4 is x to the power 5 and x into minus x cube is minus x to the power 4, x into x square is plus x cube and x into minus x is minus x square and x into plus 1 is plus x. Now multiplying by 1 to the next bracket, we get the terms x to the power 4 minus x cube plus x square minus x plus 1 equal to 0. Now cancelling the term minus x to the power 4 plus x to the power 4 plus x cube minus x cube minus x square plus x square plus x minus x. Now this polynomial is reduced to x to the power 5 plus 1 equal to 0. Now this is the required polynomial. Now taking this one to the right side, we get it as x to the power 5 is equal to minus 1. Now we have to write this minus 1 in terms of polar form and we know the polar form of minus 1 is cos pi plus i sin pi. Now pause this video and write down a formula to find out fifth root of this complex number cos pi plus i sin pi. I hope that all of you have written the answer. The formula for finding the fifth root of the complex number cos pi plus i sin pi is x is equal to cos of 2k pi plus pi divided by 5 plus i sin of 2k pi plus pi divided by 5 where k equal to 0, 1, 2, 3 and 4. Now substituting these 5 values for k, we get the required fifth roots of this polynomial x to the power 5 plus 1 equal to 0. Now taking pi common from this term, we get x is equal to cos of 2k plus 1 into pi divided by 5 plus i into sin of 2k plus 1 into pi divided by 5 where k is equal to 0, 1, 2, 3, 4. Now substituting the successive values for k, first for k equal to 0, we denote the complex number by x is 0 that is equal to putting k equal to 0 in both the terms, we get it as cos of pi by 5 plus i sin of pi by 5. Now substituting the next value of k equal to 1, we denote the complex number by x 1 that is equal to putting k equal to 1 here we get it as cos of 3 pi by 5 plus i sin of 3 pi by 5. The next value of k is equal to 2, we will denote the complex number by x 2 that is equal to putting k equal to 2 here we get it as cos of 5 pi by 5 plus i sin of 5 pi by 5. Now cancelling 5, we get it as cos pi plus i sin pi and we know that the value of cos pi is minus 1 and the value of sin pi is 0 therefore we get this root as simply minus 1. Next value of k is 3, we denote the complex number by x 3 and it is equal to after putting k equal to 3, we get it as cos of 7 pi by 5 plus i sin of 7 pi by 5 that is equal to now we know that by using the relation of cos of 2 pi minus theta as cos theta and sin of 2 pi minus theta as minus sin theta, we can write this cos of 7 pi by 5 as cos of 3 pi by 5 and this sin 7 pi by 5 as minus of sin 3 pi by 5 and this i as it is. The next value of k is 4, we will denote the complex number by x 4 that is equal to putting k equal to 4 here we get it as cos of 9 pi by 5 plus i sin 9 pi by 5 again using the above result we can write this cos 9 pi by 5 as cos of pi by 5 and sin 9 pi by 5 is nothing but minus of sin pi by 5 and this i is as it is. Now actually the given polynomial is of fourth degree and here we have obtained in all 5 roots one root extra we have obtained due to the multiplication of a linear polynomial x plus 1. Now due to the polynomial x plus 1 one root x equal to minus 1 is extra here we have obtained. So removing this root the required roots of the given polynomial are x is equal to cos of pi by 5 plus minus i sin of pi by 5 next pair is cos of 3 pi by 5 plus minus i sin of 3 pi by 5. Now let us consider one more question solve the equation solution here again we want to apply the concept of de Moivre's theorem. So either we can convert it to the form of x raise to n plus minus alpha equal to 0 or we can write it as a product of such a polynomials. So here we can see that from the first two terms it is possible to take x raise to 6 common and after taking x raise to 6 common we get the terms like inside the bracket x cube plus 8 plus and in the next two terms x cube plus 8 as it is and equal to 0. Now taking x cube plus 8 common from these two terms we get here in first bracket x cube plus 8 into in the next bracket x to the power 6 plus 1 equal to 0 now which implies that either x cube plus 8 is equal to 0 or x to the power 6 plus 1 equal to 0. Now considering the polynomials one by one consider the first polynomial x cube plus 8 equal to 0 now taking this 8 to the right hand side we get it as x cube equal to minus 8 writing minus as it is and writing 8 as 2 cube. Now taking cube root on both sides for this equation we get x is equal to actually here it is minus 2 cube is nothing but minus 1 into 2 cube. Now here we can write minus 1 to the power 1 by 3 into the cube root of 2 cube is a simple 2 that is equal to now writing 2 as it is now first we have to write down this minus 1 in terms of polar form and which is we know that cos pi plus i sin pi and the bracket raise to 1 by 3 as it is and we know the formula for finding the cube root of a complex number cos pi plus i sin pi is given by x is equal to this 2 as it is and the formula is cos of 2k plus 1 into pi divided by 3 plus i sin of 2k plus 1 into pi divided by 3 where k is equal to 0 comma 1 comma 2. Now putting k equal to 0 1 2 3 we get the 3 required roots of a polynomial x cube plus 8 equal to 0. Now substituting 1 by 1 first for k is equal to 0 we will denote the complex number by x is 0 which is equal to 2 into the bracket putting k equal to 0 here we get it as cos of pi by 3 plus i sin of pi by 3 which is equal to 2 as it is we know that cos pi by 3 is 1 upon 2 plus i into and sin pi by 3 is root 3 by 2 multiplying by 2 inside the bracket we get it as 1 plus i root 3. Now for k equal to 1 putting k equal to 1 here now we will denote this complex number by x 1 is equal to 2 into the bracket cos of 3 pi by 3 plus i sin of 3 pi by 3 that is equal to 2 into now cancelling 3 we get it as cos pi plus i sin pi that is equal to the value of cos pi is minus 1 and the value of sin pi is 0. So minus 1 into 2 we get it as minus 2 and the final value of k is 2 we will denote this complex number by x 2 is equal to 2 into the bracket putting k equal to 2 here we get it as cos of 5 pi by 3 plus i sin of 5 pi by 3 that is equal to 2 into the bracket now using the result cos of 2 pi minus theta as cos theta and sin of 2 pi minus theta as minus sin theta we can write cos of 5 pi by 3 as simple cos of pi by 3 and this plus i as it is and sin 5 pi by 3 can be written as minus of sin pi by 3 and we know that the value of cos of pi by 3 is 1 upon 2 and multiplying it by 2 we get the term as 1 and this minus i as it is and sin pi by 3 is root 3 by 2 multiplying it by 2 we get it as simply root 3. Now these are the 3 required roots of a polynomial x cube plus 8 equal to 0 now considering the next polynomial x to the power 6 plus 1 equal to 0 taking 1 to the right side we get it as x to the power 6 is equal to minus 1 writing minus 1 in polar form we get cos pi plus i sin pi now the 6th root of this complex number cos pi plus i sin pi is given by the formula x is equal to cos of 2 k plus 1 into pi divided by 6 plus i into sin of 2 k plus 1 into pi divided by 6 where k is equal to 0 comma 1 comma 2 comma 3 comma 4 comma 5. Now in the same way if we substitute the 6 values of k we get the 6 distinct roots of this polynomial x to the power 6 plus 1 equal to 0 as x is equal to root 3 by 2 plus minus i into 1 by 2 and the next pair is plus minus i and the final pair is minus root 3 by 2 plus minus i into 1 by 2.