Results 1 to 3 of 3

Math Help - complex numbers

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    3

    complex numbers

    hey i don't think this is the right thread but i couldn't find any other. can anyone help me with this question?

    Find all solutions in the form z = x +iy for real x and y to the equations

    i) z*3 = 2 + i

    ii)iz*2 +2z = i - 1

    iii) exp(z) = 1 + i

    cheers!!
    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
    For i) and ii) you can proceed as for ordinary algebraic equations. For iii) the solution is z= \ln (1+i) ...

    Kind regards

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

  3. #3
    Member
    Joined
    Aug 2009
    From
    Mumbai
    Posts
    83
    3) z = \ln (1+i).

    Taking logarithms to the base e.

    Now, 1+i = \sqrt{2}e^{i(\pi/4)} in the euler representation.

    So, using this in the above equation,

    z = \ln (1+i)= \ln (2^{1/2}e^{i(\pi/4)})

    i.e. z = \ln (2^{1/2}) +i(\pi/4)
    Last edited by mr fantastic; August 21st 2009 at 02:34 AM. Reason: Made the latex look a bit better
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 03:14 PM
  3. Replies: 2
    Last Post: February 7th 2009, 06:12 PM
  4. Replies: 1
    Last Post: May 24th 2007, 03:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 24th 2007, 12:34 AM

Search Tags


/mathhelpforum @mathhelpforum