State De Moivre's theorem and prove it for all integer values?

Question

State De Moivre's theorem and prove it for all integer values?

3 Answers

Impact of this question

20764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-03T20:59:36

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(cos theta + i sin theta)^n = cosn theta + isinntheta$

Proof by induction

Explanation:

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(cos theta + i sin theta)^n = cosn theta + isinntheta$

Prove by induction:

Basis case: $n = 1$ $n = 1$

$\Rightarrow {(cos θ + i sin θ)}^{1} = cos θ + i sin θ$ $=> ( cos theta + isin theta)^1 = cos theta + i sin theta$

Hence basis case holds

Asume true for $n = k$ $n=k$

${(cos θ + i sin θ)}^{k} = cos k θ + i sin k θ$ $(cos theta + i sin theta)^k = cosk theta + isinktheta$

Showing holds for $n = k + 1$ $n = k+1$

${(cos θ + i sin θ)}^{k + 1} = (cos k θ + i sin k θ) (cos θ + i sin θ)$ $(cos theta + i sin theta)^(k+1) = (cosk theta + isinktheta)(cos theta + isin theta)$

${(cos θ + i sin θ)}^{k + 1} =$ $(cos theta + i sin theta)^(k+1) =$

$= cos θ cos k θ + i sin θ cos k θ + i sin k θ cos θ - sin k θ sin θ$ $= costheta cosk theta + isinthetacosktheta + isinkthetacostheta - sinktheta sintheta$

$= cos θ cos k θ - sin k θ sin θ + i (sin θ cos k θ + sin k θ cos θ)$ $= costhetacosktheta - sinkthetasintheta + i ( sinthetacosktheta + sinktheta cos theta )$

$\Rightarrow = cos ((k + 1) θ) + i sin ((k + 1) θ)$ $=> = cos ( (k+1) theta ) + i sin ( (k+1) theta)$

Hence it holds true for $k + 1$ $k+1$

It holds for basis case $n = 1$ $n=1$ and $n = k + 1$ $n = k + 1$ for all $k$ $k$ that holds, hence holds for $n = 1$ $n =1$ , $n = 2$ $n =2$ , $n = 3$ $n=3$ etc...

$=> "holds" AA n in ZZ$

Answer 2 · 2018-05-03T21:00:45

$(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR$

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 $n in NN$ , which we can prove using Mathematical Induction:

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 $n=1$ in which case we have the trivial result:

$(cos theta+i sin theta)^1 = cos(1 theta) + isin(1 theta)$

So the given result is true when $n=1$ .

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when $n=m$ , for some $m in NN, m gt 1$ , in which case for this particular value of $m$ we have:

$(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 $n=m+1$

Induction Proof - Summary

So, we have shown that if the given result [A] is true for $n=m$ , then it is also true for $n=m+1$ . But we initially showed that the given result was true for $n=1$ so it must also be true for $n=2, n=3, n=4, ...$ and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for $n in NN$

Hence we have:

$(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \$ QED

Answer 3 · 2018-05-03T21:43:05

See below.

Explanation:

First let us prove $r[costheta+jsintheta]$ = $re^[jtheta]$ .

Let $z=costheta+jsintheta...........$ [1], therefore, #dz/[d theta $=$ -[sintheta-jcostheta]#

= $j^2sintheta+jcostheta$ = $j[costheta+jsintheta]$ [ since $j^2=-1]$

But $j[cos theta+jsin theta]$ = $z$ [ from ..... $[1]$ , so $dz/[d theta$ = $jz$ ......... $[2]$
Inverting both sides of ..... $[2]$ , $d theta/[dz$ = $1/[jz$ and integrating w.r.t $z$

$theta=1/jint1/zdz$ = $1/jlnz+c$ , but from ..... $[1]$ when $theta =0,cos theta=1,sin theta=0, i.e., z=1$

Therefore, $theta=1/jln[1]+c$ , but $ln [1] =0$ to any base, therefore $c=0$ .

So we have, $theta=1/jlnz$ , i.e, $jtheta=lnx$ , and so $e^j theta=z$ [ theory of logs], therefore from ... $[1]$ , $r[cos theta+jsin theta]=re^j theta$ ....... $[3]$

[ Proof of De Moivres theorem, for any value of $n$ ]

From ..... $[3]$ , $cos theta+jsin theta=e^j theta$ [ omitting $r$ ]....... $[4]$

Raising both sides of .... $[3]$ to the power $n$ , $[cos theta+jsin theta]^n$ = $e^[n j theta]$

= $e^[j[n theta]]$ , but $e^[j[n theta]]$ = $cosn theta+ jsin n theta$ , from ....... $[4]$ , replacing $theta$ by $n theta$ ,

Therefore , $[cos theta+jsin theta]^n$ = $cosn theta+jsin ntheta$ .

Science

Math

Humanities

... and beyond

State De Moivre's theorem and prove it for all integer values?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question