1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expressing f(z) in terms of z

  1. Jul 13, 2014 #1
    Hello! I am stuck at the final step. How do I get x+iy from the equation? Help!

    I am so sorry for posting this question in a picture instead of writing it out, because I don't know how to write equations on here.
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    I see a small mistake: on the 4th line, shouldn't you be taking out the common factor 1/4?

    Check your exponential equivalents for sine, cosine, etc., on line 3. There's your mistake.
     
    Last edited: Jul 13, 2014
  4. Jul 13, 2014 #3
    why 1/4 ?
     
  5. Jul 13, 2014 #4

    Mentallic

    User Avatar
    Homework Helper

    You substituted incorrectly.

    [tex]\cos{z}=\frac{e^{iz}+e^{-iz}}{2}[/tex]

    [tex]\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}[/tex]

    And you should factor out a value on the 4th line because you want your expression to be of the form

    [tex]k\left(\frac{e^{iz}-e^{-iz}}{2i}\right)[/tex]
    (or some similar form in the brackets)
    where k is some complex number, and z is a complex number. Your expression would then simplify to [itex]k\sin{z}[/itex]
     
  6. Jul 13, 2014 #5
    But z=x+iy. How do I change i(x^2-y^2) into z? I can't figure out how to rearrange it.
     
  7. Jul 13, 2014 #6

    Mentallic

    User Avatar
    Homework Helper

    Well of course you should recognize that

    [tex]z^2=(x+iy)^2=x^2-y^2+i 2xy[/tex]

    and what you have is VERY similar to this form. Since you made a mistake early on that I pointed out, your final result isn't going to work with the method I'm hinting at here, but when you fix that up then it'll fall into place.

    [tex]i(x^2-y^2)-2xy[/tex]

    Should be quite easily converted into a function of z by observing the z2 result.
     
  8. Jul 14, 2014 #7
    Ok, so I have corrected my mistakes but I got stuck here. How do I continue?
     

    Attached Files:

  9. Jul 14, 2014 #8

    Mentallic

    User Avatar
    Homework Helper

    Now expand like you did earlier and simplify where possible.
     
  10. Jul 14, 2014 #9
    I did that but still couldn't get the answer. Hold on, I will show you my work later.
     
  11. Jul 14, 2014 #10

    Mentallic

    User Avatar
    Homework Helper

    Did you end up with

    [tex]-i\left(\frac{e^{w}-e^{-w}}{2}\right)[/tex]

    where

    [tex]w=i(x^2-y^2)-2xy[/tex]

    ??
     
  12. Jul 14, 2014 #11
    I don't know what to do next. How do I simplify this further?
     
    Last edited: Jul 14, 2014
  13. Jul 14, 2014 #12
    Sorry, did a mistake earlier. Here's what I got.
     

    Attached Files:

  14. Jul 14, 2014 #13

    Mentallic

    User Avatar
    Homework Helper

    Sorry, I only just glossed over your previous upload and didn't spot the error.

    [tex]\sinh{z}=\frac{e^z-e^{-z}}{2}[/tex]

    while you had an i in the denominator.
     
  15. Jul 14, 2014 #14
    Oh yes, that was my mistake. But then what happens to the i outside of the bracket?
     
  16. Jul 14, 2014 #15

    Mentallic

    User Avatar
    Homework Helper

    The first 4 terms have a constant factor of

    [tex]\frac{1}{4i}[/tex]

    while the right 4 are

    [tex]\frac{i}{4}[/tex]

    How do these two numbers relate to each other?
     
  17. Jul 14, 2014 #16
    They're minus to each other. Sorry, I don't know how to say that mathematically.
     
  18. Jul 14, 2014 #17

    Mentallic

    User Avatar
    Homework Helper

    Right!

    [tex]\frac{1}{4i}=\frac{i}{4(-1)}=\frac{-i}{4}[/tex]

    and so now you can continue to simplify the expression!
     
  19. Jul 14, 2014 #18
    Simplified it, then what? I am still stuck here.
     

    Attached Files:

  20. Jul 14, 2014 #19

    Mentallic

    User Avatar
    Homework Helper

    Check post #10.
     
  21. Jul 14, 2014 #20
    Uhmm, how do I change w into z? Can I just simpy do this?

    exp(i*(ia-b)) ?? Is it possible to multiply i just like that?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted