 Hello and welcome to the session. In this session, we will discuss a question which says that using DeMauroves theorem, find all the values of 2 minus 2 iota by root 3, 4 raised to power 1 by 4, also finding the continued product of the three values. Now before starting the solution of this question, we should know a result. And that is DeMauroves theorem. According to this, if n is any integer positive or negative, then s theta plus iota sine theta whole raised to power n is equal to cos n theta sine n theta. Any fraction positive or negative, then plus iota sine theta whole raised to power n is equal to cos n theta. What kind of the key idea for solving this question? And now we will start with the solution. We have to find all the values of this, all the values of 2 minus 2 iota by root 3 whole raised to power 1 by 4, cos theta root 3 is equal to r sine theta. And let us name it as 1 and this as now an adding of minus is equal to r square cos square theta times square theta is equal to r square into cos square theta square theta by root 2 by 3 is equal to into, now cos square theta plus sine square theta is equal to 1 and r square into 1 is equal to r square. This figure applies to 16 by 3 which gave to 4 by root 3. Now here we will consider the positive value, this value or modulus of the complex number r s theta cos theta root 3 into cos theta and this implies cos theta is equal to root 3 by 2. Question number 2 minus 2 by root 3 is equal to r minus 1 by 2. So we have is equal to root 3 by 2 is equal to minus 1 by 2. Now since theta minus theta is negative therefore theta minus theta is equal to over root tan theta is equal to minus 1 by root 3 is equal to minus 10. Now this in fact is equal to minus pi by 6 which further than that from these two we have theta is equal to 5 pi by 6. These two we have theta is equal to 11 pi by 6 theta is in the second quadrant and this value of theta is in the fourth quadrant. Now we have discussed earlier that theta minus 11 pi by 6 in the fourth quadrant into iota will be equal to 1 by 4 is equal to theta into iota will raise to pi 1 by 4. Now taking the common it will be equal to r raise to pi 1 by 4 into now cos theta. Now theta here is 11 pi by 6 so this will be cos into iota whole raise to pi 1 by 4 to 4 by root 3 whole raise to pi 1 by 4 into now here by using the result that is here is equal to cos theta as theta is equal to sin theta written as raise to pi 1 by 4 which is given in the key idea is 1 by 4 and this is theta raise to pi 1 by 4 into which is 1 by 4 theta which is 2 and pi plus 11 pi by 6 which is 1 by 4 theta which is 11 pi by 6. 2 raise to pi 1 by 4 into pi by 24 plus iota sin is equal to 0 1 root 3 into iota whole raise to pi 1 by 4 so 1 and 2 will find which will become 4 by root 3 whole raise to pi 1 by 4 into cos 0 plus 11 pi by 24 plus iota into sin 0 plus 11 pi by 24 the whole which will be further equal to 2 power 1 by 4 into cos 11 pi by 24 by by 24 whole raise to pi 1 by 4 and pi by 24 the whole expression will be 4 by root 3 whole raise to pi 1 by 4 into cos 0 by 4 plus 11 pi by 24 plus 11 pi by 24 4 by root 3 whole raise to pi 1 by 4 into cos 23 pi by 24 23 pi by 24 the whole raise to pi 1 by 4 into cos 23 pi by 24 plus iota by 24 the whole. Now for iota sin 11 pi by 24 plus iota sin 11 pi by 24 the whole which is equal to 4 by root 3 whole raise to pi 1 by 4 into pi by 24 plus iota sin 35 pi by 24 the whole therefore into cos 35 pi by 24 plus iota sin by the continued product of these three values will be 4 by root 3 whole raise to pi 1 by 4 into cos by 24 plus 23 pi by 24 plus 35 pi by 24 plus iota sin by 24 plus 23 pi by 24 plus 35 pi by 24 the whole power 1 by 4 into on solving this it will be 5 by 24 and on solving this it will be 10 equal to whole raise to pi 1 by 4 into cos on solving this it will be 23 pi by 8 plus iota sin 23 pi by 8 23 pi by 8 and sin 23 pi by 8 we are getting minus cos pi by 8 pi by 8 respect to root 3 1 raise to pi 1 by 4 cos pi by 8 4 into cos pi by 8 minus iota sin pi by so this is the solution of the given question and that's all for the session hope you all have enjoyed the session