 Hello everyone, myself A.S. Falmari, Assistant Professor, Department of Humanities and Sciences, Valchan Institute of Technology, Solaapur. In the last video, we have discussed D. Merver's Theorem and few examples. In this video also, we consider some examples on how to evaluate the powers of a complex numbers. The learning outcome is, at the end of this session, students will be able to evaluate powers of complex numbers using D. Merver's Theorem. Now, consider the first example. Evaluate 1 plus sin alpha plus i cos alpha divided by 1 plus sin alpha minus i cos alpha bracket power n. Solution, now here we use one trick that we write 1 as sin square alpha plus cos square alpha and in this one, we replace this plus by minus i square. We know that i square is the value minus 1, so minus minus will nothing but again plus. Now, writing this plus as minus of i square. Now applying the result of a square minus b square is equal to a minus b into a plus b. We get it as sin alpha plus i cos alpha into sin alpha minus i cos alpha. To get this numerator, now let us add sin alpha plus i cos alpha on both sides. After adding, we get it as 1 plus sin alpha plus i cos alpha is equal to sin alpha plus i cos alpha plus in the first bracket sin alpha plus i cos alpha into sin alpha minus i cos alpha. Now, here we can see that it is possible to take sin alpha plus i cos alpha common from these two terms. After taking this term common, we get sin alpha plus i cos alpha inside the bracket into, now the first term left as a simple 1 plus from the second one, the left quantity is sin alpha minus i cos alpha. Now, in order to have this given quantity, we can see that this is what a denominator. Let us divide this whole equation by 1 plus sin alpha minus i cos alpha, we get 1 plus sin alpha plus i into cos alpha divided by 1 plus sin alpha minus i into cos alpha is equal to, now this will be removed, we are left with only sin alpha plus i cos alpha. Now, here we have to take the nth power of this quantity, but if I observe on the right hand side, here we have to definitely apply de Moivre's theorem, but the first term is sin 1. So, we convert it to cos by using the relation between them sin n cos like sin alpha is equal to cos of pi by 2 minus alpha plus i into and writing this cos alpha as sin of pi by 2 minus alpha. Now, taking nth power on the both sides, we get 1 plus sin alpha plus i cos alpha divided by 1 plus sin alpha minus i cos alpha bracket raise to n is equal to on the right hand side, this quantity is present inside the bracket cos of pi by 2 minus alpha plus i sin of pi by 2 minus alpha bracket raise to n. Now, this is what a structure of a de Moivre's theorem, let us apply de Moivre's theorem. Before applying de Moivre's theorem, we get it as cos of n into the bracket pi by 2 minus alpha plus i sin of n into the bracket pi by 2 minus alpha. This is the simplified form of this complex number. Let us consider one more example, evaluate cos theta plus i sin theta divided by sin theta plus i cos theta bracket raise to 4 solution. The given expression is cos theta plus i sin theta divided by sin theta plus i cos theta bracket raise to 4 taking the separate power of numerator and denominator, we get it as cos theta plus i sin theta bracket raise to 4 divided by sin theta plus i cos theta bracket raise to 4. Let us call it as equation number one and evaluating numerator and denominator separately. Consider first numerator cos theta plus i sin theta bracket raise to 4 is equal to immediately by application of de Moivre's theorem, we get it as cos 4 theta plus i sin 4 theta. Consider the denominator sin theta plus i cos theta bracket raise to 4. Let us write down this sin theta as cos of pi by 2 minus theta plus i into and writing cos theta as sin of pi by 2 minus theta bracket raise to 4. Now by applying de Moivre's theorem, we get cos of 4 into the bracket pi by 2 minus theta plus i into sin of 4 into the bracket pi by 2 minus theta. Now after multiplying by 4 inside the bracket, we get it as cos of 2 pi minus 4 theta plus i into sin of inside the bracket 2 pi minus 4 theta. Now pause this video and simplify these two terms. I hope that all of you have simplified it. Simplifying these two terms is nothing but applying the trigonometric identity cos of A minus B which is nothing but cos A cos B plus sin A sin B and sin of A minus B for this term which is nothing but sin A into cos B minus cos A into sin B. Now applying this trigonometric result for these two terms, we get cos 2 pi into cos 4 theta plus sin 2 pi into sin 4 theta plus i into inside the bracket sin 2 pi into cos 4 theta minus cos 2 pi into sin 4 theta. Now the value of cos 2 pi is 1 so 1 into cos 4 theta is cos 4 theta. We know that the value of sin 2 pi is 0 therefore second term vanishes plus i into the bracket. Now the value of sin 2 pi is 0 therefore this term is also vanishes and this is minus the value of cos 2 pi is 1 so it is minus sin 4 theta and this i as it is. Now applying the corollary number 2 here we can write this minus as a power minus 1 for the term cos 4 theta plus i sin 4 theta bracket raise to minus 1. Now substituting the expression for numerator and denominator in the equation number 1 we get now this is what the equation number 1 substituting the expressions we get it as cos 4 theta plus i sin 4 theta divided by cos 4 theta plus i sin 4 theta bracket raise to minus 1. In the denominator this quantity is present with negative power means we have to write it in the numerator with positive sign therefore we get this product both the terms are same with 1 1 as a power we can write it as cos 4 theta plus i sin 4 theta bracket raise to 2. Now again applying DeMauver's theorem we get cos of 8 theta plus i sin 8 theta. Let us consider one more example if x plus 1 upon x is equal to 2 cos theta then prove that 2 cos r theta is equal to x to the power r plus 1 upon x to the power r solution. Now we have a given relation is x plus 1 upon x is equal to 2 cos theta so in order to solve such a kind of an equation so every time our aim is to convert it to a polynomial in order to convert it to a polynomial we multiply it by x we get it as x square plus 1 upon x into x is 1 is equal to 2 x cos theta rewriting it we get x square minus 2 x cos theta plus 1 is equal to 0. Now we can consider it as a quadratic polynomial in x and it has 2 roots the roots of this quadratic polynomial are obtained by using the formula x is equal to minus b plus minus under root of b square minus 4 ac upon 2 a here a is 1 b is minus 2 cos theta and c is 1. Now substituting all the values of a b c in this formula we get it as x is equal to minus into the bracket minus 2 cos theta plus minus under root of minus 2 cos theta bracket square minus 4 into 1 into 1 upon 2 into 1. Now simplifying this we get it as now this minus minus will become plus 2 cos theta plus minus under root of minus 2 cos theta bracket square is nothing but 4 cos square theta minus 4 into 1 into 1 is simple 4 divided by 2 into 1 is 2. Now from this under roots and we can take 4 outside and under root of 4 is 2 again from this we can take 2 common and cancel with the denominator so we are left with only cos theta plus minus under root of cos square theta minus 1. Again writing cos theta as it is plus minus under root of we know that cos square theta minus 1 is nothing but minus of sin square theta it is equal to cos theta as it is plus minus we know that under root of minus sin square theta is nothing but i sin theta. Now it has two roots one is cos theta plus i sin theta and another is cos theta minus i sin theta. Now let us consider the positive root x is equal to cos theta plus i sin theta. Now taking rth power on both sides we get x to the power r is equal to cos theta plus i sin theta bracket raise to r. Applying de Malvers theorem we get it as cos r theta plus i sin r theta call it as equation number 1. Consider the quantity 1 upon x to the power r which is nothing but x to the power minus r and x is nothing but cos theta plus i sin theta and bracket raise to minus r. Now applying de Malvers theorem directly we get it as cos of r theta due to the minus and we get here minus i sin of r theta. Now let us call it as equation number 2. In order to calculate the given quantity we have to add equation number 1 and 2. After adding we get it as x to the power r plus 1 upon x to the power r is equal to cos r theta plus i sin r theta plus in the next bracket cos r theta minus i sin r theta. Now we can see that the second and fourth one are same terms with opposite sign. We have to remove them and the first one and the third term is the same one. So we can write it as 2 times cos r theta hence proved.