Results 1 to 12 of 12

Math Help - Complex equation

  1. #1
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24

    Complex equation

    Hi I have this equation

    (e^{\pi\cdot z})^{2}=i

    Where I have to find the roots, when modulus er less than 1.

    Cannot really solve it...

    I have tried to t=e^{\pi\cdot z}

    And then tried to solve it is a quacdric equation.

    Where I get that

    e^{\pi x}\cos(\pi y)=1 and e^{\pi x}\sin(\pi y)=-2

    And also for the other root

    e^{\pi x}\cos(\pi y)=-1 and e^{\pi x}\sin(\pi y)=2

    But I have that e^{\pi x}< 1 \Leftrightarrow x<0

    So anyone who could help me out or give me some tips?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by lo2 View Post
    Hi I have this equation

    (e^{\pi\cdot z})^{2}=i

    Where I have to find the roots, when modulus er less than 1.

    Cannot really solve it...

    I have tried to t=e^{\pi\cdot z}

    And then tried to solve it is a quacdric equation.

    Where I get that

    e^{\pi x}\cos(\pi y)=1 and e^{\pi x}\sin(\pi y)=-2

    And also for the other root

    e^{\pi x}\cos(\pi y)=-1 and e^{\pi x}\sin(\pi y)=2

    But I have that e^{\pi x}< 1 \Leftrightarrow x<0

    So anyone who could help me out or give me some tips?
    e^{2 \pi z} = e^{i\left(\frac{\pi}{2} + 2 n \pi\right)}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24
    But is modulus not then 1 or more. Because x=0 and then we have.

    e^{0}(\cos(y)+i\sin(y))

    And thus modulus is 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    following on from Mr F post.

    2 \pi z = i\left(\frac{\pi}{2} + 2 n \pi\right)
     z = i\left(\frac{1+4n}{4}\right)
     |z| = \left|\left(\frac{1+4n}{4}\right) \right|

    for what values of n is the modulus of that fraction less than 1 ?

    Bobak
    Follow Math Help Forum on Facebook and Google+

  5. #5
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24
    I am sorry but I do not even really understand the first calculation. How can
    (e^{\pi z})^{2}=e^{i(\frac{\pi}{2}+2n\pi)} And what is n and what happens to x.

    Me trying:
    (e^{\pi z})^{2}=e^{2\pi z}=e^{2\pi (x+iy)}=e^{(2\pi x+2\pi iy)}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by lo2 View Post
    I am sorry but I do not even really understand the first calculation. How can
    (e^{\pi z})^{2}=e^{i(\frac{\pi}{2}+2n\pi)} And what is n and what happens to x.

    Me trying:
    (e^{\pi z})^{2}=e^{2\pi z}=e^{2\pi (x+iy)}=e^{(2\pi x+2\pi iy)}
    Index Law: (a^m)^n = a^{mn}.

    i = e^{i \frac{\pi}{2}} = e^{i \left(\frac{\pi}{2} + 2 n \pi\right)}. n is an integer: n = 0, \, \pm 1, \, \pm 2, \, ....

    Why introduce z = x + iy etc. when you can solve for z directly?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24
    Now I have finally managed to solve it and I get that.

    z=i(\frac{1}{2}+2)

    If I put in p=0 I get that z=i \frac{1}{4}

    If I put p=1 then modulus is more than 1

    But If I put in p=-1 then I get a modulus that is not too high and I get this solution z=i -\frac{3}{4}

    But I am just not sure that it is in fact a solution to this equation. So is it?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    z = \frac{1}{4}i \ \ , \ \ - \frac{3}{4}i are all the solutions.

    However I think you made a few typos in your pervious post.

    Bobak
    Follow Math Help Forum on Facebook and Google+

  9. #9
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24
    yeah...

    But thank you very much! Your help is most appreciated!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,615
    Thanks
    1578
    Awards
    1
    Are you looking for solutions? I don't follow this thread.
    \left( {e^{\frac{{\pi i}}<br />
{4}} } \right)^2  = i\,\& \,\left( {e^{\frac{{ - 3\pi i}}<br />
{4}} } \right)^2  = i
    Follow Math Help Forum on Facebook and Google+

  11. #11
    lo2
    lo2 is offline
    Newbie
    Joined
    Aug 2006
    Posts
    24
    I just wanted to ask whether z=-\frac{3}{4}i was a solution to the equation (e^{z \pi})^2=i
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,615
    Thanks
    1578
    Awards
    1
    Quote Originally Posted by lo2 View Post
    I just wanted to ask whether z=-\frac{3}{4}i was a solution to the equation (e^{z \pi})^2=i
    Well I just wanted to ask you what \Large {e^{\left( {\frac{{ - 3\pi i}}{4}} \right)}} =?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex equation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 20th 2010, 10:22 PM
  2. Complex equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 22nd 2010, 02:31 AM
  3. complex equation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 29th 2010, 02:38 AM
  4. Complex equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 21st 2008, 03:38 AM
  5. A complex equation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 8th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum