 Hello everyone, myself AS Falmari, Assistant Professor, Department of Humanities and Sciences, Balchan Institute of Technology, Solapur. Students, in this video, we are going to learn D-Moevers' Theorem. Learning outcome, at the end of this session, students will be able to evaluate the powers of complex numbers using D-Moevers' Theorem. Let us start this video with the theorem. D-Moevers' Theorem is, for any rational number n, the value or one of the value of cos theta plus i sin theta bracket raised to n is equal to cos n theta plus i sin n theta. D-Moevers' Theorem has the many applications. It is used to evaluate the powers of complex numbers. It is also used to find out the roots of a complex number, roots of an equation. So, in this video, we will see how to find out the powers of certain complex numbers using D-Moevers' Theorem. Before we start the example, it has some corollaries. Corollary number one is that cos theta minus i sin theta bracket raised to n is equal to cos n theta minus i sin n theta. So, the difference between the D-Moevers' Theorem and the corollaries, the middle sign in the corollary is minus and for the D-Moevers' Theorem is plus. Still, for such an expression, it is possible to write down this power as a coefficient in the angles for cos and sin. How it is possible? Let us see. Now, the given expression is cos theta minus i into sin theta bracket raised to n. Now, our aim is to convert it to the standard form of a D-Moevers' Theorem cos of theta plus i sin theta. In order to write here, now we will absorb this minus n sin using the property of a sin which is an odd function. And we can write this minus sin theta as sin of minus theta and outside we can write plus i to have the same angles for cos n sin and cos is an even function. So, cos of minus theta is again cos theta so that we can replace this cos theta as cos of minus theta here. Now, here we can see that cos of same angle plus i sin of same angle and having brackets some rational number. Now, we can apply D-Moevers' Theorem here. We get it as cos of minus n theta plus i sin of minus n theta. Now, again using the even and odd property of the cos and sin, we get it as cos of minus n theta equal to cos n theta. Cos is an even function and sin is a non-function therefore sin of minus n theta is minus of sin n theta and this i is as it is. One more corollary cos theta minus i sin theta bracket raised to n is equal to cos theta plus i sin theta bracket raised to minus n. Sometimes it is required to write down such a kind of a quantity in terms of cos theta plus i sin theta. In such a case, we use this corollary number 2. So, let us see how it is true. So, given quantity is cos theta minus i sin theta bracket raised to n. As we have seen above, we can absorb this minus n sin by writing like this cos of minus theta plus i sin of minus theta. Now, here we can see that in the angles for cos and sin, minus 1 coefficient is present. So, applying the Moevers' Theorem in the reverse way that is writing this minus 1 coefficient in the power. We get it as minus n and the bracket we have cos theta plus i sin theta. Some important notes while applying the Moevers' Theorem. To apply the Moevers' Theorem, the real part must be the cosine 1 and the imaginary part must be the sin 1. That is, the Moevers' Theorem cannot be directly applied to sin theta plus i cos theta bracket raised to n. Here, this is not a standard structure, but it is possible to convert it to the structure of a Moevers' Theorem. The procedure is what? This is what a given thing sin theta plus i cos theta bracket raised to n. We know the relation between cos and sin like this. Sin theta can be converted into cos by the result sin theta equal to cos of pi by 2 minus theta and cos theta can be converted to sin of pi by 2 minus theta. Now, we can see that now it is what the standard structure of a Moevers' Theorem and some bracket power is present. Now, we can apply the Moevers' Theorem. We get it as cos of n into the bracket pi by 2 minus theta plus i sin n into pi by 2 minus theta. So, this is what the procedure when the real part is given as a sin and imaginary part as a cos. But in such a case, we cannot apply the Moevers' Theorem directly. One more important notice, to apply the Moevers' Theorem, the angles present with sin and cos are must be same. That is, the Moevers' Theorem cannot be applied to cos theta plus i sin pi bracket raised to n because the structure is same, but the angles present with cos and sin are different. So, in such a case, it is not possible to use the Moevers' Theorem. These two results are very important. So, let us consider examples. Simplify cos 5 theta minus i sin 5 theta bracket raised to 2 into cos 7 theta plus i sin 7 theta bracket raised to minus 3 divided by cos 4 theta minus i sin 4 theta bracket raised to 9 into cos theta plus i sin theta bracket raised to 5. This is what a very big expression. Now, we have to convert it to a simple complex number form. To simplify such a complex number, we consider each term and we are trying to write it in the form of cos theta plus i sin theta with some power. Now, consider the first term cos 5 theta minus i sin 5 theta bracket raised to 2. Now, according to corollary number 2, we can take this 5 and minus in the power so that the term left is cos theta plus i sin theta bracket raised to minus 5 into 2 is minus 10. Consider cos 7 theta plus i sin 7 theta bracket raised to minus 3. Now, we are applying the Moevers' Theorem in reverse order. We get it as cos theta plus i sin theta bracket raised to this 7. We have to write in the power, but in the power already minus 3 is present. So, 7 into minus 3 is minus 21. Now, consider the next bracket cos 4 theta minus i sin 4 theta bracket raised to 9. Applying corollary number 2 here, we can write it as cos 4 theta plus i sin 4 theta and taking this minus sign in the power, we get it as minus 9. Now, applying the Moevers' Theorem that is writing this 4 in the power, we get it as cos theta plus i sin theta bracket raised to 4 into minus 9, which is minus 36. And the next bracket, which is also in a standard form. Now, let us substitute all these expressions for these terms in this given quantity. After substituting, we get this expression. Now, here we can see that the brackets are same. We can add and subtract the powers. Now, the first power is minus 10. Next one is minus 21. So, it is minus 10, minus 21 is minus 31. In the denominator, it is minus 36 and 5, which is again minus 31. Both the terms are same, so that after cancelling them, we get the value as 1. So, this big expression, which looks like very complex in nature, it is nothing but a simple number 1. Pause this video and find sin theta minus i cos theta bracket raised to 6. I hope that all of you have written the answer. Now, the problem is to find out the value of sin theta minus i cos theta bracket raised to 6. Now, the solution. Now, we can see that the structure is not like a Moevers' Theorem. So, we use this relation to convert sin to cos and cos to sin. Now, writing sin theta as cos of pi by 2 minus theta and minus i as it is and writing cos theta as sin of pi by 2 minus theta and the power bracket 6 as it is. Now, let us apply the Moevers' Theorem directly. We get it as what? Cos of 6 into pi by 2 minus theta minus i sin of 6 into pi by 2 minus theta. That is equal to multiplying by this 6 inside the bracket. We get it as cos of 6 pi by 2 is 3 pi minus 6 theta minus i sin of again multiplying by this 6 into the same bracket. We get it as 3 pi minus 6 theta. Now, in order to further simplification, let us use the result of cos of a minus b here. The result is cos a cos b plus sin a sin b. We get it as cos 3 pi cos 6 theta plus sin 3 pi sin 6 theta minus i and applying here the result sin of a minus b which is nothing but sin a into cos b minus cos a into sin b and after applying we get it as sin 3 pi into cos 6 theta minus sin 6 theta into cos 3 pi. Now, we know the value of cos 3 pi is minus 1. So, the first term is minus cos 6 theta and the value of sin 3 pi is 0. Therefore, the second term vanishes minus i as it is inside the bracket. Again here sin 3 pi is present. Again its value is 0. Therefore, we can write it as 0 plus 0 and this minus and again cos 3 pi is present and cos 3 pi is minus 1. So, this minus minus will become plus and sin 6 theta as it is. Then finally, we get it as minus cos 6 theta minus i sin 6 theta.