Results 1 to 7 of 7
Like Tree2Thanks
  • 1 Post By Prove It
  • 1 Post By Plato

Math Help - Complex equation

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    sweden
    Posts
    61

    Complex equation

    Hi everybody

    I need some help to solve this complex equation:


    (√3 + i − ez) (ez-1) = 0


    that has a modulus smaller than 3π.

    I'm thinking about using zero product property:

    ez - 1 = 0

    z = ln(1)

    z = 0

    but when it comes to the first parenthesis I don't know how to proceed.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,658
    Thanks
    1480

    Re: Complex equation

    Quote Originally Posted by Tala View Post
    Hi everybody

    I need some help to solve this complex equation:


    (√3 + i − ez) (ez-1) = 0


    that has a modulus smaller than 3π.

    I'm thinking about using zero product property:

    ez - 1 = 0

    z = ln(1)

    z = 0

    but when it comes to the first parenthesis I don't know how to proceed.

    Thank you.
    If we remember that \displaystyle \begin{align*} z = x + i\,y \end{align*} with \displaystyle \begin{align*} x, y \in \mathbf{R} \end{align*}, setting the first factor equal to 0 gives

    \displaystyle \begin{align*} \sqrt{3} + i - e^z &= 0 \\ \sqrt{3} + i &= e^z \\ 2e^{\frac{\pi}{6} i} &= e^{x + i\,y} \\ 2e^{\frac{\pi}{6} i} &= e^x e^{i\,y} \\ e^x = 2 \textrm{ and } e^{\frac{\pi}{6} i } &= e^{i\,y} \\ x = \ln{2} \textrm{ and } y &= \frac{\pi}{6} \end{align*}

    Therefore \displaystyle \begin{align*} z = \ln{2} + \frac{\pi}{6} i \end{align*}
    Thanks from Tala
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    sweden
    Posts
    61

    Re: Complex equation

    This might be a stupid question but I can't see how the argument is π/6 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2012
    From
    sweden
    Posts
    61

    Re: Complex equation

    I get it know:

    arccos(√(3)/2) right ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,658
    Thanks
    1480

    Re: Complex equation

    Quote Originally Posted by Tala View Post
    This might be a stupid question but I can't see how the argument is π/6 ?
    If the complex number is in the first quadrant, which it is, \displaystyle \begin{align*} \textrm{arg}\,(z) = \arctan{\left[ \frac{\textrm{Im}\,(z)}{\textrm{Re}\,(z)} \right]} \end{align*}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2012
    From
    sweden
    Posts
    61

    Re: Complex equation

    Thank you so much for the quick and brilliant answers.
    I really appreciate it.
    Tak (it means thank you in danish).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1

    Re: Complex equation

    Quote Originally Posted by Prove It View Post
    Therefore \displaystyle \begin{align*} z = \ln{2} + \frac{\pi}{6} i \end{align*}
    It is worth noting that for each k\in\mathbb{Z},~\ln (2) + i\left( {\frac{\pi }{6} + 2k\pi } \right) is also a solution.
    Thanks from Prove It
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex equation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 26th 2010, 05:24 AM
  2. Complex equation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 20th 2010, 10:22 PM
  3. complex equation
    Posted in the Algebra Forum
    Replies: 7
    Last Post: February 9th 2009, 10:45 AM
  4. Complex equation
    Posted in the Calculus Forum
    Replies: 11
    Last Post: October 1st 2008, 03:42 PM
  5. Complex equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 21st 2008, 03:38 AM

Search Tags


/mathhelpforum @mathhelpforum