Results 1 to 4 of 4

Thread: Solve e^z - 2 = 2i.

  1. #1
    Junior Member
    Mar 2011

    Solve e^z - 2 = 2i.

    Hi, could someone help me with this one?

    Find all solutions of $\displaystyle e^{z}-2 = 2i$

    Last edited by mr fantastic; Mar 23rd 2011 at 03:40 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    Define $\displaystyle \displaystyle z = x + iy$.

    $\displaystyle \displaystyle e^z - 2 = 2i$

    $\displaystyle \displaystyle e^z = 2 + 2i$.

    Working on the RHS:

    $\displaystyle \displaystyle 2 + 2i = 2(1 + i)$

    $\displaystyle \displaystyle = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}$, where $\displaystyle \displaystyle n \in \mathbf{Z}$.

    Working on the LHS:

    $\displaystyle \displaystyle e^z = e^{x + iy} = e^x\,\textrm{cis}\,y$.

    So $\displaystyle \displaystyle e^x\,\textrm{cis}\,y = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}$.

    Clearly $\displaystyle \displaystyle e^x = 2\sqrt{2} \implies x = \ln{(2\sqrt{2})}$ and $\displaystyle \displaystyle y = \frac{\pi}{4} + 2\pi n$.

    Therefore, $\displaystyle \displaystyle z = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right)$.

    Alternatively, just use the fact that $\displaystyle \displaystyle \log{Z} = \ln{|Z|} + i\arg{Z}$.

    If $\displaystyle \displaystyle e^z = 2(1 + i) $

    Then $\displaystyle \displaystyle z = \log{\left[2(1 + i)\right]}$

    $\displaystyle \displaystyle = \ln{|2(1 + i)|} + i\arg{[2(1 + i)]}$

    $\displaystyle \displaystyle = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right)}$ where $\displaystyle \displaystyle n \in \mathbf{Z}$.
    Last edited by Prove It; Mar 22nd 2011 at 04:41 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Apr 2005
    To start with, do the same thing you would if it were a "real numbers" problem. Add 2 to both sides to get
    $\displaystyle e^z= 2+ 2i$
    and then $\displaystyle z= ln(2+ 2i)$

    Now, if a complex number, z, is written in "polar form", $\displaystyle z= re^{i\theta}$ then $\displaystyle ln(z)= ln(re^{i\theta)}= ln(r)+ i\theta$

    So just write 2+ 2i in "polar form". Don't forget that whatever $\displaystyle \theta$ is, adding $\displaystyle 2\pi$ to it doesn't change z but will give different logarithms. That the reason for the "all solutions".
    Last edited by HallsofIvy; Mar 23rd 2011 at 05:20 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Mar 2011
    Thanks so much, appreciate it!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Jul 16th 2010, 10:29 PM
  2. Replies: 1
    Last Post: Jun 9th 2009, 10:37 PM
  3. how do I solve this?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jan 22nd 2009, 06:21 PM
  4. How could i solve?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 2nd 2009, 02:18 PM
  5. Solve for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jan 1st 2009, 12:33 PM

Search tags for this page

Search Tags

/mathhelpforum @mathhelpforum