A community for students.
Here's the question you clicked on:
 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
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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2oh oh oh ok ok i think i got something! a clever little trick. lemme explain it and see if this is working.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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:\[\large\rm e^{iz}e^{iz}=2z\qquad\qquad(1)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \color{orangered}{\sin(z)}=z\]Alternatively, since sine and cosine are cofunctions, we can also establish this:\[\large\rm \color{orangered}{\cos\left(\frac{\pi}{2}z\right)}=z\]And taking the same route as before,\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{i\left(\frac{\pi}{2}z\right)}+e^{i\left(\frac{\pi}{2}z\right)}\right)}=z\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And then you have to do a little calculation at that point:\[\large\rm e^{i(\pi/2z)}=e^{i \pi/2}\cdot e^{z}=i\cdot e^{z}\]Ok with that step?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, the sin(z) would have a 2i in its exponential form.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Woops*\[\large\rm e^{i(\pi/2z)}=e^{i \pi/2}\cdot e^{iz}=i\cdot e^{iz}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oo good call, I missed that in the denominator.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm e^{iz}e^{iz}=2iz\qquad\qquad(1)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do you understand where I'm going with this other equation though? :o\[\large\rm \color{orangered}{\frac{1}{2}\left(i e^{iz}i e^{iz}\right)}=z\]\[\large\rm e^{iz}e^{iz}=\frac{2z}{i}\] \[\large\rm e^{iz}e^{iz}=2iz\qquad\qquad (2)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I guess I should've expected that :) lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aww. 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Another 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,\[\large\rm (e^{iz})^21=2iz(e^{iz})\]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,\[\large\rm w^22izw1=0\]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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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: \(x = \pm \sqrt{\cosh^{2}y  y^{2}\coth^{2}y}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not allowed to use that theorem. The idea behind this is to show it algebraically.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can it be?? sinz =z > sin z  z=0 iff z = 2kpi?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, because then you would have \(\sin(2k\pi) 2k\pi = 0 \implies 2k\pi = 0\)
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.