State De Moivre's theorem and prove it for all integer values?
3 Answers
Proof by induction
Explanation:
Prove by induction:
Basis case:
Hence basis case holds
Asume true for
Showing holds for
Hence it holds true for
It holds for basis case
# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR#
Explanation:
De Moivre's Theorem, states that:
# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR#
We seek to validate this result for
Induction Proof - Hypothesis
We seek to prove that:
#
# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \ # ..... [A]
So let us test this assertion, [A], using Mathematical Induction:
Induction Proof - Base case:
Consider, the special case
# (cos theta+i sin theta)^1 = cos(1 theta) + isin(1 theta) #
So the given result is true when
Induction Proof - General Case
Now, Let us assume that the given result [A] is true when
# (cos theta+i sin theta)^m = cos(m theta) + isin(m theta) \ \ \ # ..... [B]
Then, consider the expression:
# E = (cos theta+i sin theta)^(m+1) #
# \ \ \ = (cos theta+i sin theta)^m(cos theta+i sin theta) #
# \ \ \ = (cos(m theta) + isin(m theta))(cos theta+i sin theta) \ \ # (using [B])
# \ \ \ = cos(m theta)cos theta + isin(m theta)cos theta + cos(m theta)i sin theta + isin(m theta)i sin theta #
# \ \ \ = {cos(m theta)cos theta -sin(m theta) sin theta} + i{sin(m theta)cos theta + icos(m theta) sin theta } #
# \ \ \ = cos(m+1) theta + isin(m+1) theta #
Which is the given result [A] with
Induction Proof - Summary
So, we have shown that if the given result [A] is true for
Induction Proof - Conclusion
Then, by the process of mathematical induction the given result [A] is true for
Hence we have:
# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \ # QED
See below.
Explanation:
First let us prove
Let
=
But
Inverting both sides of .....
Therefore,
So we have,
[ Proof of De Moivres theorem, for any value of
From .....
Raising both sides of ....
=
Therefore ,