A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

How many solutions exist in the complex plane to sin(z) = z? Justify your answer. Note: Meant to be shown algebraically somehow, not with theorems

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's the justification that I have no idea how to show. I know there are infinitely many solutions, but just not sure quite what to do. I know there are theorems that could help, but they are not theorems we have touched on, so this is something intended to be done algebraically. I tried setting up a system of equations with it, but I then just ran into a dead end mess. I don't think series could help but maybe? Anyone have any ideas on an algebraic approach to show this?

  2. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh oh oh ok ok i think i got something! a clever little trick. lemme explain it and see if this is working.

  3. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\large\rm \color{orangered}{\sin(z)}=z\]\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{iz}-e^{-iz}\right)}=z\]So this will give us our first equation:\[\large\rm e^{iz}-e^{-iz}=2z\qquad\qquad(1)\]

  4. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\large\rm \color{orangered}{\sin(z)}=z\]Alternatively, since sine and cosine are co-functions, we can also establish this:\[\large\rm \color{orangered}{\cos\left(\frac{\pi}{2}-z\right)}=z\]And taking the same route as before,\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{i\left(\frac{\pi}{2}-z\right)}+e^{-i\left(\frac{\pi}{2}-z\right)}\right)}=z\]

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    And then you have to do a little calculation at that point:\[\large\rm e^{i(\pi/2-z)}=e^{i \pi/2}\cdot e^{-z}=i\cdot e^{-z}\]Ok with that step?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well, the sin(z) would have a 2i in its exponential form.

  7. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Woops*\[\large\rm e^{i(\pi/2-z)}=e^{i \pi/2}\cdot e^{-iz}=i\cdot e^{-iz}\]

  8. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oo good call, I missed that in the denominator.

  9. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\large\rm e^{iz}-e^{-iz}=2iz\qquad\qquad(1)\]

  10. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Do you understand where I'm going with this other equation though? :o\[\large\rm \color{orangered}{\frac{1}{2}\left(i e^{-iz}-i e^{iz}\right)}=z\]\[\large\rm e^{-iz}-e^{iz}=\frac{2z}{i}\] \[\large\rm e^{iz}-e^{-iz}=2iz\qquad\qquad (2)\]

  11. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh darn it! :( Now it's just giving us 0=0. I forgot about that i on the sine, and it was giving me z=0 with that mistake lol.

  12. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I guess I should've expected that :) lol

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Aww. That was clever, though, the cofunction identity. Didn't consider that, lol. Yeah, I was trying to mess around with: \(\sin(x+iy) = x+iy\) \(\sin(x)\cosh(y) + i\sinh(y)\cos(x) = x+iy\) \(\sin(x)\cosh(y) = x\) and \(\sinh(y)\cos(x) = y\) And then yeah, seeing what I could do by multiplying or dividing, possibly using the pythagorean identities, etc. Just wasnt really getting anywhere with it.

  14. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Another idea I had was uhhh, so that identity from before, we get to this point,\[\large\rm e^{iz}-e^{-iz}=2iz\]Multiplying through by e^{iz} gives us,\[\large\rm (e^{iz})^2-1=2iz(e^{iz})\]And then I was trying to make a sort of substitution so we could deal with a simple quadratic,\[\large\rm w=e^{iz}\]giving us,\[\large\rm w^2-2izw-1=0\]But that other z is causing a problem :P grr. Hmm I might be out of ideas :( I think I need some sleep anyway lol

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No worries. I like the cofunction identity attempt, though. Maybe I can mess around with someone identities I wasn't considering. I know there's a way to do this, haha.

  16. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    .

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure if I should be going down the route of taking a system and then trying to get a result like 1 = 1 like might happen in a basic system of equations. That doesn't seem like something that is legit, though. Not that itd help, but I did manage to get this at some point: \(x = \pm \sqrt{\cosh^{2}y - y^{2}\coth^{2}y}\)

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not allowed to use that theorem. The idea behind this is to show it algebraically.

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can it be?? sinz =z --> sin z - z=0 iff z = 2kpi?

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, because then you would have \(\sin(2k\pi)- 2k\pi = 0 \implies 2k\pi = 0\)

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.