Page 2 of 2 FirstFirst 12
Results 16 to 19 of 19
Like Tree10Thanks

Thread: Solve using de Moivre's theorem: z^3=i

  1. #16
    Newbie
    Joined
    Jan 2015
    From
    Martin
    Posts
    3

    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Prove It View Post
    First of all, when dealing with solving exponential equations, the EXPONENTIAL form of the complex number is ALWAYS easiest to use...

    $\displaystyle \begin{align*} z^3 &= \mathrm{i} \\ z^3 &= \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{3}} \\ z &= \mathrm{ e }^{ \frac{\pi}{6} + \frac{2\pi\,n}{3} } \end{align*}$

    So in the first cycle, the possible solutions are $\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}}, \mathrm{e}^{\frac{5\pi}{6}} , \mathrm{e}^{\frac{3\pi}{2}} \end{align*}$, which in Cartesian form are:

    $\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*}$

    I am really bad in this. Sorry guys. I cant really solve it. Can you try solve whole example and like for real bad student ?
    For example how did you get from i e on something ?
    If someone have time would be nice if he can solve this examples but every step what you do..
    I have like 4-5 examples and cant really solve it. and i have exam in 2 days
    Best would be way like" Prove it " solved it. We are learning it like that.
    z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
    or is there some manual where they describe everything.
    Anyway thats for help what you allready solved.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,738
    Thanks
    1896

    Re: Solve using de Moivre's theorem: z^3=i

    If you can put your equation in "cis" form, then you can put the equation in exponential form. In general $\displaystyle \begin{align*} r\,\mathrm{e}^{\mathrm{i}\,\theta} \equiv r \left[ \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \right] \end{align*}$
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor

    Joined
    Aug 2006
    Posts
    20,488
    Thanks
    2329
    Awards
    1

    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Martinun View Post
    For example how did you get from i e on something ?
    If someone have time would be nice if he can solve this examples but every step what you do..
    I have like 4-5 examples and cant really solve it. and i have exam in 2 days
    Best would be way like" Prove it " solved it. We are learning it like that.
    z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
    You are a victim of poor mathematical education practice. There is a good trend to use a more geometric approach to teaching complex analysis.

    First you need to understand that any complex has n nth roots. Those n roots are located on a circle centered at the origin of radius the nth root of absolute value of the number. Those n roots divide the circle into n equal arcs.

    Let's stop a moment for an example. There are six sixth roots of 3-4\bf{i}. They are located on a circle centered at (0,0) and radius \sqrt[6]5. On that circle the roots are separated by arcs of \dfrac{2\pi}{6}, where 2\pi is the whole circle and six equal parts.

    Now give some notation to simplify the way we can list the roots.
    Let \exp(i\theta)=\cos(\theta)+i\sin(\theta).
    Then for a complex number z=x+yi then \exp(z)=e^x\left(\exp(iy)\right)=\sqrt{x^2+y^2}\bf  {cis}(\theta).
    Finally we give the notation for the argument of the complex number:
    Suppose that x\cdot y\ne 0 then \theta= Arg(x + yi) = \left\{ {\begin{array}{{lr}}  {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\   {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\   {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.

    Here is a example of the process: z^5=2-2i. \bf{Arg}(2-2i})=\theta=\arctan\left(\frac{-2}{2}\right)
    This is the principle root, \rho=\sqrt[5]{|2-2i|}\exp\left(\dfrac{i\theta}{5}\right), so that \rho^5=2-2i

    List all the roots. Let \zeta=\exp\left(\frac{2\pi}{5}~\bf{i}\right)
    the five roots are: \rho\cdot\zeta^k,~k=0,1,2,3,4
    Last edited by Plato; Jan 19th 2015 at 07:37 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,576
    Thanks
    2590

    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Martinun View Post
    Can you please tell me how you get 4k+1 from 2kpi ?
    MarkFL didn't get 4k+ 1 from 2k\pi.

    What he had was \frac{\pi}{2}+ 2k\pi. Getting the "common denominator", that is \frac{\pi}{2}+ \frac{4k\pi}{2} = \frac{\pi+ 4k\pi}{2}= \frac{\pi}{2}(4k+ 1).
    Last edited by HallsofIvy; Feb 27th 2015 at 09:18 AM.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Using De Moivre's Theorem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Mar 28th 2011, 02:32 AM
  2. De Moivre's Theorem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 18th 2010, 01:43 PM
  3. De Moivre Theorem Help
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 24th 2009, 10:51 PM
  4. Regarding De Moivre’s Theorem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 10th 2009, 12:23 PM
  5. De Moivre's Theorem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 28th 2008, 08:59 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum