Results 1 to 3 of 3

Math Help - de Moivre's theorem

  1. #1
    Junior Member
    Joined
    Jan 2011
    From
    Sydney
    Posts
    36

    de Moivre's theorem

    Hi, I'm currently trying to deduce de Moivre's formula,

    (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}.

    This is what i have so far...

    If z = \cos{\theta} + i\sin{\theta}, \arg{z} = \theta and \arg{z^n} = n\theta

    Then we have,

    z^n = r^n(\cos{\theta}+i\sin{\theta})^n\\= (\cos{\theta}+i\sin{\theta})^n\\=\cos(\arg(z^n))+i  \sin(arg(z^n))\\=\cos{n\theta} + i\sin{n\theta}

    But I can't seem to figure out how they got from the second to the third step, It would be greatly appreciated if someone could please explain this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: de Moivre's theorem

    The Euler's identity...

    e^{i \theta} = \cos \theta + i\ \sin \theta (1)

    ... combined with elementary properties of exponential conducts immediately to the result...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428

    Re: de Moivre's theorem

    Quote Originally Posted by liedora View Post
    Hi, I'm currently trying to deduce de Moivre's formula,

    (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}.

    This is what i have so far...

    If z = \cos{\theta} + i\sin{\theta}, \arg{z} = \theta and \arg{z^n} = n\theta

    Then we have,

    z^n = r^n(\cos{\theta}+i\sin{\theta})^n\\= (\cos{\theta}+i\sin{\theta})^n\\=\cos(\arg(z^n))+i  \sin(arg(z^n))\\=\cos{n\theta} + i\sin{n\theta}

    But I can't seem to figure out how they got from the second to the third step, It would be greatly appreciated if someone could please explain this.
    You can use induction...

    You wish to show \displaystyle (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}.

    Base step: \displaystyle n = 2.

    \displaystyle \begin{align*} (\cos{\theta} + i\sin{\theta})^2 &= \cos^2{\theta} + 2i\sin{\theta}\cos{\theta} + i^2\sin^2{\theta} \\ &= \cos^2{\theta} - \sin^2{\theta} + 2i\sin{\theta}\cos{\theta} \\ &= \cos{2\theta} + i\sin{2\theta} \end{align*}

    Inductive step: Assume the statement is true for \displaystyle n = k, in other words, assume \displaystyle (\cos{\theta} + i\sin{\theta})^k = \cos{k\theta} + i\sin{k\theta}, then we need to show the statement is true for \displaystyle n = k + 1, in other words, that \displaystyle (\cos{\theta} + i\sin{\theta})^{k+1} = \cos{(k + 1)\theta} + i\sin{(k + 1)\theta}.

    \displaystyle \begin{align*}(\cos{\theta} + i\sin{\theta})^{k + 1} &= (\cos{\theta} + i\sin{\theta})^k(\cos{\theta} + i\sin{\theta}) \\ &= (\cos{k\theta} + i\sin{k\theta})(\cos{\theta} + i\sin{\theta}) \\ &= \cos{k\theta}\cos{\theta} + i\cos{k\theta}\sin{\theta} + i\cos{\theta}\sin{k\theta} + i^2\sin{k\theta}\sin{\theta} \\ &= \cos{k\theta}\cos{\theta} - \sin{k\theta}\sin{\theta} + i(\cos{k\theta}\sin{\theta} + \cos{\theta}\sin{k\theta}) \\ &= \cos{(k\theta + \theta)} + i\sin{(k\theta + \theta)}\\ &= \cos{(k + 1)\theta} + i\sin{(k + 1)\theta} \end{align*}

    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using De Moivre's Theorem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 15th 2011, 04:15 PM
  2. De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: August 31st 2009, 05:45 AM
  3. De Moivre's Theorem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 15th 2009, 01:43 PM
  4. Regarding De Moivre’s Theorem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 10th 2009, 11:23 AM
  5. De Moivre's theorem.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 2nd 2009, 07:23 PM

Search Tags


/mathhelpforum @mathhelpforum