Results 1 to 3 of 3
Like Tree4Thanks
  • 2 Post By Idea
  • 2 Post By HallsofIvy

Thread: solve complex equation

  1. #1
    Member
    Joined
    May 2009
    Posts
    243
    Thanks
    3

    solve complex equation

    Hi folks,

    I am trying to solve the complex equation $z^4 + 4 = 0$

    I did it by setting $z = a + i b$ and the results came out. 1 + i, 1 - i, -1 + i, -1 - i.

    I then tried to do the same thing putting z in polar form and have no such luck.

    $(\cos t + i \sin t)^4 + 4 = 0$ gives

    $ \cos 4t + 4 + i \sin 4t = 0$ using de Moivre's theorem.

    so $ \cos 4t + 4 = 0 $ ........ real

    and $ \sin 4t = 0 $ ........ imaginary

    the imaginary part gives $4t = n \pi$ i.e. $t = \dfrac{n\pi}{4} $

    when $n = 0$, $z = 1$

    when $n = 1$, $z = \dfrac{1}{\sqrt{2}} + i\dfrac{1}{\sqrt{2}}$

    when $n = 2$, $z = i$

    the real part gives

    $\cos 4t + 4 = 0$

    it can be shown that $\cos 4t = 8\cos^4 t- 8\cos^2 t + 1$

    $8\cos ^4 t - 8 \cos^2 t + 5 = 0$ let $x = \cos^2 t$

    $8x^2 - 8x + 5 = 0$

    $x = \dfrac{4 \pm i \sqrt{14}}{8} $ and this looks nothing like the correct answer.

    I can't see any mistakes in my work, but there is no way this is getting to the right answer.

    Can anyone see what I am doing wrong?
    Last edited by s_ingram; Mar 31st 2018 at 10:28 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    854
    Thanks
    402

    Re: solve complex equation

    $\left|z|^4=4\right.$

    $|z|=\sqrt{2}$

    $z=\sqrt{2}(\cos t + i \sin t)$
    Thanks from topsquark and s_ingram
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,924
    Thanks
    3097

    Re: solve complex equation

    Your error was in writing z as cos(t)+ i sin(t). That is true only for z having absolute value 1. But, as Idea points out, in order that z satisfy $\displaystyle z^4= 4$, z must have absolute value $\displaystyle \sqrt[4]{4}= \sqrt{2}$.

    You could write z as rcos(t)+ ri sin(t) for real r and t.
    Thanks from topsquark and s_ingram
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 14th 2012, 08:32 PM
  2. solve the equation (complex number)
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Sep 1st 2012, 12:06 PM
  3. Complex numbers, solve equation
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: Mar 12th 2011, 03:20 PM
  4. Complex equation to solve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 22nd 2009, 12:42 AM
  5. Replies: 1
    Last Post: Oct 28th 2007, 08:44 AM

/mathhelpforum @mathhelpforum