Results 1 to 12 of 12
Like Tree2Thanks
  • 1 Post By Deveno
  • 1 Post By Soroban

Math Help - De Moivre's Theorem

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    localhost
    Posts
    5

    Question [Solved] De Moivre's Theorem

    Please help me solve this problem.

    if a=cos2alpha + isin2alpha
    b=cos2beta + isin2beta
    prove that,

    a-b / a+b = itan (alpha-beta)
    Last edited by ok59; August 20th 2012 at 07:44 AM. Reason: solved
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: De Moivre's Theorem

    Assuming "2alpha" is 2 \alpha, then

    a = e^{2i \alpha}
    b = e^{2i \beta}

    Then, \frac{a-b}{a+b} = \frac{e^{2i \alpha} - e^{2i \beta}}{e^{2i \alpha} + e^{2i \beta}}. Can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    localhost
    Posts
    5

    Re: De Moivre's Theorem

    sorry i dont get it ( still learning )
    how do i suppose to get tan from it.
    do you mean (cos alpha + isin alpha)^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: De Moivre's Theorem

    Quote Originally Posted by ok59 View Post
    sorry i dont get it ( still learning )
    how do i suppose to get tan from it.
    do you mean (cos alpha + isin alpha)^2
    Yes, it's also equal to (\cos \alpha + i \sin \alpha)^2 by de Moivre's. I don't know if that helps though. Expanding doesn't yield anything useful.

    It could also help to play around with the RHS, i \tan (\alpha - \beta). Note that this is a purely imaginary number. You can try to reduce the LHS as i \frac{\sin (\alpha - \beta)}{\cos (\alpha - \beta)} or something like that. You'll most probably need your trigonometric identities.
    Last edited by richard1234; August 19th 2012 at 10:07 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2012
    From
    localhost
    Posts
    5

    Re: De Moivre's Theorem

    hmm.. this problem is so freaking weird . I am about to give up ( it has been 2 days).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: De Moivre's Theorem

    Yeah, it's an interesting problem. I haven't solved it yet...there seem to be several possible ways to the solution though. Just try a bunch of methods, see if one of them works. Maybe another user has a nice, elegant solution that doesn't require a lot of brute force (I don't want to brute force algebra/trig identities right now...). There might even be a geometric solution involving two points on a circle and finding the tangent of the angle in between.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: De Moivre's Theorem

    let's write, for the time being, cos(2α) = u, sin(2α) = v, cos(2β) = y, sin(2β) = z.

    then:

    \frac{a-b}{a+b} = \frac{u-y + i(v-z)}{u+y + i(v+z)} = \frac{((u-y) + i(v-z))((u+y) - i(v+z))}{(u+y)^2 + (v+z)^2}

    = \frac{(u-y)(u+y) + (v-z)(v+z) + i(-(u-y)(v+z) + (v-z)(u+y))}{u^2 + 2uy + y^2 + v^2 + 2vz + z^2}

    =\frac{u^2 + v^2 - y^2 - z^2 + i(2(vy - uz))}{u^2 + v^2 + y^2 + z^2 + 2(uy + vz)}

    now we can use some trig to simplify.

    note that u2+v2 = y2+z2 = 1. so our messy fraction becomes:

    \frac{1 - 1 + i(2(vy - uz))}{2 + 2(uy + vz))} = i\frac{vy - uz}{1 + uy + vz}

    now vy - uz = \sin(2\alpha)\cos(2\beta) - \cos(2\alpha)\sin(2\beta) = \sin(2\alpha - 2\beta)

    and uy + vz = \cos(2\alpha)\cos(2\beta) + \sin(2\alpha)\sin(2\beta) = \cos(2\alpha - 2\beta).

    so what we have is:

    i\frac{\sin(2\alpha - 2\beta)}{1 + \cos(2\alpha - 2\beta)} = i\tan\left(\frac{2\alpha - 2\beta}{2}\right) = i\tan(\alpha - \beta)
    Thanks from ok59
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2012
    From
    localhost
    Posts
    5

    Re: De Moivre's Theorem

    Man ! Thank so very much Deveno.
    You made it so simple to understand mate. You can't imagine the time i've spent on this problem.
    Thank You again for your help.
    Have a nice day !

    edit: Thank you to Richard1234 too. I have solved it by your way too. Now I have two solutions.
    Last edited by ok59; August 20th 2012 at 07:39 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744

    Re: De Moivre's Theorem (Part 1)

    Hello, ok59!

    Out of curiosity, I brute-forced it . . . whew!

    We need the Sum-to-Product Identities:

    . . \sin\!A\;+\;\sin\!B \:=\: 2\sin\!\left(\tfrac{A+B}{2}\right)\cos\!\left( \tfrac{A-B}{2}\right) \qquad \cos\!A\;+\;\cos\!B \:=\: 2\cos\!\left(\tfrac{A+B}{2}\right)\cos\!\left( \tfrac{A-B}{2}\right)

    . . \sin\!A\;-\;\sin\!B \:=\:2\cos\!\left(\tfrac{A+B}{2}\right)\sin\!\left  ( \tfrac{A-B}{2}\right) \qquad \cos\!A\;-\;\cos\!B \:=\:\text{-}2\sin\!\left(\tfrac{A+B}{2}\right)\sin\!\left( \tfrac{A-B}{2}\right)


    \text{Given: }\:\begin{Bmatrix}a &=& \cos2A + i\sin2A \\ b &=& \cos2B + i\sin2B \end{Bmatrix}

    \text{Prove that: }\:\frac{a-b}{a+b} \:=\; i\tan(A-B)

    \text{The numerator is: }\:a - b \:=\:(\cos2A-\cos2B) + i(\sin2A - \sin 2B)

    x x . . . . . . . . . . . . . . . =\;\text{-}2\sin(A\!+\!B)\sin(A\!-\!B) + 2i\cos(A\!+\!B)\sin(A\!-\!B)

    x x . . . . . . . . . . . . . . . =\;2\sin(A\!-\!B)\bigg[\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)\bigg]


    \text{The denominator is:  }\:a + b \:=\:(\cos2A + \cos2B) + i(\sin2A +\sin2B)

    . . . . . . . . . . . . . . . . . . . =\;2\cos(A\!+\!B)\cos(A\!-\!B) + 2i\sin(A\!+\!B)\cos(A\!-\!B)

    . . . . . . . . . . . . . . . . . . . =\;2\cos(A\!-\!B)\bigg[\cos(A\!+\!B) + i\sin(A\!+\!B)\bigg]


    \text{And we have: }\:\frac{a-b}{a+b} \;=\;\frac{\sin(A\!-\!B)\bigg[\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)\bigg]}{\cos(A\!-\!B)\bigg[\cos(A\!+\!B) + i\sin(A\!+\!B)\bigg]}

    . . . . . . . . . . . . . . . =\;\tan(A\!-\!B)\left[\frac{\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)}{\cos(A\!+\!B) + i\sin(A\!+\!B)\right]}


    \text{Rationalize: }\:\tan(A-B)\cdot\left[\frac{\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)}{\cos(A\!+\!B) + i\sin(A\!+\!B)}\right]\cdot\left[\frac{\cos(A\!+\!B) - i\sin(A\!+\!B)}{\cos(A\!+\!B) - i\sin(A\!+\!B)}\right]

    . . . . . . . =\;\tan(A-B)\cdot \left[\frac{\text{-}\sin(A\!+\!B)\cos(A\!+\!B) + i\sin^2(A\!+\!B) + i\cos^2(A\!+\!B) - i^2\sin(A\!+\!B)\cos(A\!+\!B)} {\cos^2(A\!+\!B) - i^2\sin^2(A\!+\!B)}\right]

    . . . . . . . =\;\tan(A-B)\cdot\left[\frac{\text{-}\sin(A\!+\!B)\cos(A\!+\!B) + i\sin^2(A\!+\!B) + i\cos^2(A\!+\!B) + \sin(A\!+\!B)\cos(A\!+\!B)}{\cos^2(A\!+\!B) + \sin^2(A\!+\!B)}\right]

    . . . . . . . =\;\tan(A-B)\cdot\left[\frac{i\sin^2(A+B) + i\cos^2(A+B)}{1}\right]

    . . . . . . . =\;\tan(A-B)\cdot i\left[\sin^2(A+B) + \cos^2(A+B)\right]

    . . . . . . . =\;\tan(A-B)\cdot i \cdot 1

    . . . . . . . =\;i\tan(A-B)
    Last edited by Soroban; August 20th 2012 at 09:30 AM.
    Thanks from ok59
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2012
    From
    localhost
    Posts
    5

    Re: De Moivre's Theorem (Part 1)

    Thanks Mr. Soroban for another method . Now I understand that the trick is to divide and multiply with the conjugate of denominator to get the result. I appreciate the efforts of the members of this board.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jul 2012
    From
    jaipur, Rajasthan,India
    Posts
    24
    Thanks
    1

    Re: De Moivre's Theorem

    De Moivre's Theorem is a relatively simple formula for calculating powers of complex numbers.De Moivre's formula states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).(r cisθ)^n = r^n cis(nθ) here n is an integer.
    free algebra word problems
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: [Solved] De Moivre's Theorem

    Here's another (equivalent) way: I'll begin with a general calculation that I'll then apply to this specific problem.

    Suppose z \in \mathbb{C} \ni z \neq -1, and |z|^2 = 1 (i.e. z on the unit circle except the leftmost point).

    So let z = e^{\theta i}, where \theta is not an odd multiple of \pi.

    Then \frac{z-1}{z+1}

    = \frac{(z-1)(\bar{z}+1)}{(z+1)(\bar{z}+1)} = \frac{|z|^2+z-\bar{z}-1}{|z|^2+z+\bar{z}+1} = \frac{1+z-\bar{z}-1}{1+z+\bar{z}+1}

    = \frac{z-\bar{z}}{2+(z+\bar{z})} = \frac{2Im(z)i}{2+2Re(z)} = \frac{Im(z)}{1+Re(z)}i = \frac{\sin(\theta)}{1+\cos(\theta)}i = \tan(\theta / 2)i.

    With that in hand, the result follows quickly. Let a=e^{2\alpha i}, b=e^{2\beta i}, a+b \neq 0. Let z = a/b. (Note that |b| = 1, so b \neq 0.)

    Let \theta = 2(\alpha - \beta). Then z = e^{\theta i}, and obviously |z| = 1.

    Notice that from a+b \neq 0, it follows that a \neq -b, so that z = a/b \neq -1, and hence so can apply the above derivation with this z.

    Now consider \frac{a-b}{a+b}. Have:

    \frac{a-b}{a+b} = \frac{a/b - 1}{a/b + 1} = \frac{z-1}{z+1} = \tan(\theta / 2)i = \tan((2(\alpha - \beta)) / 2)i = \tan(\alpha - \beta)i as desired.

    Thus the claim is established.

    Notice that all the "bad" cases amounted to a + b = 0, because then a = -b = e^{(2n+1) \pi i}b, so e^{2\alpha i} = e^{(2n+1) \pi i}e^{2\beta i}, so \theta = 2(\alpha - \beta) = (2n+1) \pi.

    That would've made z = e^{\theta i} = e^{(2n+1) \pi i} = -1, and also would've maded \alpha - \beta = (n+1/2) \pi, which is where \tan(\alpha - \beta)i is undefined.
    Last edited by johnsomeone; September 13th 2012 at 03:47 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. de Moivre's theorem
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: July 17th 2011, 04:31 AM
  2. [SOLVED] De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: November 18th 2010, 02:06 PM
  3. [SOLVED] De Moivre's theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 30th 2010, 08:53 AM
  4. De Moivre's Theorem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 28th 2008, 08:59 AM
  5. De Moivre's theorem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 17th 2008, 01:01 AM

Search Tags


/mathhelpforum @mathhelpforum