 Hello everyone. Myself AS Phalmari Assistant Professor, Department of Humanities and Sciences, Valchan Institute of Technology, Solapur. In this video we consider few more different types of examples for finding the roots. Learning outcome at the end of this session, students will be able to find the roots of a complex number using de Moivre's theorem. Now let us consider the first example. Find all the values of 1 upon 2 plus i into root 3 by 2 whole bracket raise to 3 by 4 and show that their continued product is 1. Now the solution. Now here we have given the complex number in Cartesian form. So our first step is to express this Cartesian form complex number into polar form. We know that any complex number x plus i y can be written in polar form as r into the bracket cos theta plus i sin theta where we have to determine two quantities r and theta and that are given by r is equal to under root of x square plus y square and theta is equal to tan inverse of y by x. Comparing given complex number with respect to number x plus i y, we get the value of x as 1 by 2 and y as root 3 by 2. Now let us calculate r. r is equal to under root of the formula is given here x square here x is 1 by 2 so that we can write here 1 upon 2 bracket square plus and the next quantity is y square here y is root 3 by 2 therefore we write here root 3 by 2 bracket square is equal to under root of 1 by 2 square is 1 by 4 plus and root 3 by 2 square is square of root 3 is simple 3 divided by square of 2 is 4 is equal to under root of now here the denominator is same we can write it as it is 4 now the numerators are 1 and 3 that is 1 plus 3 is 4 and under root of 4 by 4 we can cancel 4 from the numerator and denominator we get it as root of 1 and root of 1 is a simple 1. Next calculating theta, theta is given as theta is equal to tan inverse of y divided by x that is equal to tan inverse of here y is given as root 3 by 2 divided by and x is given as 1 divided by 2. Now writing it now we can see that this 2 is in the denominator and this 2 is nothing but in the numerator after rewriting it so we can remove this 2 we get it as tan inverse of root 3 and we know that tan inverse of root 3 is nothing but pi by 3. Therefore we have obtained the values of r and theta we can write down the given complex number 1 upon 2 plus i into root 3 by 2 in terms of polar form as here h is r into but the value of r is 1 therefore it is 1 into the bracket cos of theta here we have obtained as pi by 3 plus i sin pi by 3. Now we have to take the 3 by 4 power of this complex number so let us take 3 by 4 power on both sides we get 1 by 2 plus i into root 3 by 2 bracket raised to 3 by 4 is equal to again taking the 3 by 4 power on the right hand side we get it as cos of pi by 3 plus i into sin of pi by 3 bracket raised to 3 by 4. Now actually we want to write here the power somewhat like 1 upon n in order to write it in terms of 1 upon n I can write this numerator power inside like this this is equal to internal bracket cos of pi by 3 plus i sin pi by 3 this internal bracket raised to 3 and whole bracket raised to 1 by 4 now first we will apply de Moivre's theorem for the internal bracket we get cos of 3 pi by 3 plus i sin of 3 pi by 3 and whole bracket raised to 1 by 4 as it is we can cancel these three from cos and sin and we get it as cos of pi plus i sin pi bracket raised to 1 by 4 this is nothing but the fourth root of a complex number cos pi plus i sin pi we know that the nth root of any complex number cos theta plus i sin theta is given as cos of 2k pi plus theta upon n plus i sin 2k pi plus theta upon n now here theta is pi and n is 4 therefore we get z is equal to cos of 2k pi plus pi upon 4 plus i sin of 2k pi plus pi upon 4 here a is equal to 0 comma 1 comma 2 comma 3 now this equation will provide us all the required values of given complex number finding them one by one now from this let us take pi common we can write z as z is equal to cos of after taking pi common we get inside the bracket 2k plus 1 into pi upon 4 plus i into sin of 2k plus 1 into pi upon 4 for k is equal to 0 now substituting k is equal to 0 and denoting the value as z 0 we get if we put k is 0 so it is 2 into 0 is 0 0 plus 1 is 1 into pi that is simply we get it as pi by 4 therefore we can write z 0 is equal to cos of pi by 4 plus i sin of pi by 4 we know that the value of cos pi by 4 is 1 upon root 2 plus i into again the value of sin of pi by 4 is 1 upon root 2 the next value of k is 1 we denote that complex number by z 1 and is equal to cos of 3 pi by 4 plus i into sin of 3 pi by 4 and it is equal to cos of now we can write this 3 pi by 4 as pi minus pi by 4 plus i sin pi minus pi by 4 and we know that cos of pi minus theta is nothing but minus of cos theta using this relation we get here minus cos of theta is pi by 4 plus i sin of again applying the result sin of pi minus theta as simple sin theta we get here sin of pi by 4 is equal to now writing minus as it is value of cos of pi by 4 is 1 upon root 2 plus i and value of sin of pi by 4 is 1 upon root 2 the next value of k is 2 z 2 and it is equal to cos of 5 pi by 4 plus i sin of 5 pi by 4 now using the result cos of pi plus theta as minus cos theta and sin of pi plus theta as minus sin theta we get it as z 2 is equal to minus cos pi by 4 minus i into sin of pi by 4 that is equal to writing minus as it is value of cos of pi by 4 is 1 upon root 2 again writing this minus i as it is and the value of sin of pi by 4 is 1 upon root 2 and the last value of k is 3 we will denote the value of z as z 3 that is equal to putting k equal to 3 here we get 2 into 3 is 6 6 plus 1 is 7 so we get it as 7 pi by 4 now here we get z 3 is equal to cos of 7 pi by 4 plus i sin of 7 pi by 4 is equal to now writing this 7 pi by 4 as cos of 2 pi minus pi by 4 plus i sin of 2 pi minus pi by 4 again using the result cos of 2 pi minus theta is simple cos theta and sin of 2 pi minus theta as minus sin theta we get here z 3 is equal to cos of pi by 4 minus i into sin of pi by 4 that is equal to value of cos pi by 4 is 1 upon root 2 and this minus i as it is and value of sin pi by 4 is 1 upon root 2 now these are the z 0 z 1 z 2 z 3 are the all the values of given complex number 1 upon 2 plus i root 3 by 2 whole power 3 by 4 now the next problem is we have to show that if we take the continuous product of all these values z 0 z 1 z 2 z 3 so it must be 1 it is we have to show that now consider their continuous product z 0 into z 1 into z 2 into z 3 is equal to z 0 here we have obtained as 1 upon root 2 plus i into 1 upon root 2 into and z 1 we have obtained as minus 1 upon root 2 plus i into 1 upon root 2 into and z 2 we have obtained as minus 1 upon root 2 minus i into 1 upon root 2 and finally z 3 we have obtained as 1 upon root 2 minus i into 1 upon root 2 now here we can see that a complex number present in the first bracket and a complex number present in the last bracket both are the complex conjugates of each other and we know that the product of a complex conjugates is nothing but the sum of the squares of real and imaginary part so real part of this complex number is 1 upon root 2 and imaginary part is 1 upon root 2 so the product of these two complex number we can write as 1 upon root 2 square plus 1 upon root 2 square into again we can see that the complex number present in the second bracket and third bracket are also complex conjugates of each other again using the same result we can write its value as square of real part and here the real part is minus 1 upon root 2 plus square of imaginary part the imaginary part is 1 upon root 2 and that is equal to square of 1 upon root 2 is 1 upon 2 plus again here it is square of 1 upon root 2 we can write 1 upon 2 into and again here it is square of minus 1 upon root 2 which is nothing but 1 upon 2 plus again it is square of 1 upon root 2 is 1 upon 2 now for in the first bracket 1 upon 2 plus 1 upon 2 is 1 into again in the second bracket it is 1 upon 2 plus 1 upon 2 which is also 1 and 1 into 1 is 1 now here we have shown that the continuous product of all the values of a given complex number is 1 pause this video and write a complex number 1 plus i in polar form i hope all of you have written the answer the problem is to write down a complex number 1 plus i in the polar form and we know that any complex number which is present in the Cartesian form like x plus i y can be written in polar form as r into the bracket cos theta plus i sin theta where r is given as under root of x square plus y square and theta is given as tan inverse of y by x now calculating r r is equal to under root of x square plus y square is equal to under root of 1 square plus 1 square that is equal to root 2 and theta is nothing but tan inverse of y by x is equal to tan inverse of y is 1 divided by x is also 1 that is equal to tan inverse of 1 and we know that tan inverse of 1 is pi by 4 therefore the given complex number 1 plus i can be written in polar form as root 2 into the bracket cos of pi by 4 plus i sin of pi by 4