anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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?
zepdrix
  • zepdrix
oh oh oh ok ok i think i got something! a clever little trick. lemme explain it and see if this is working.
zepdrix
  • zepdrix
\[\large\rm \color{orangered}{\sin(z)}=z\]\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{iz}-e^{-iz}\right)}=z\]So this will give us our first equation:\[\large\rm e^{iz}-e^{-iz}=2z\qquad\qquad(1)\]

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zepdrix
  • zepdrix
\[\large\rm \color{orangered}{\sin(z)}=z\]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,\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{i\left(\frac{\pi}{2}-z\right)}+e^{-i\left(\frac{\pi}{2}-z\right)}\right)}=z\]
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
Well, the sin(z) would have a 2i in its exponential form.
zepdrix
  • zepdrix
Woops*\[\large\rm e^{i(\pi/2-z)}=e^{i \pi/2}\cdot e^{-iz}=i\cdot e^{-iz}\]
zepdrix
  • zepdrix
Oo good call, I missed that in the denominator.
zepdrix
  • zepdrix
\[\large\rm e^{iz}-e^{-iz}=2iz\qquad\qquad(1)\]
zepdrix
  • zepdrix
Do you understand where I'm going with this other equation though? :o\[\large\rm \color{orangered}{\frac{1}{2}\left(i e^{-iz}-i e^{iz}\right)}=z\]\[\large\rm e^{-iz}-e^{iz}=\frac{2z}{i}\] \[\large\rm e^{iz}-e^{-iz}=2iz\qquad\qquad (2)\]
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
I guess I should've expected that :) lol
anonymous
  • anonymous
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.
zepdrix
  • zepdrix
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
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
.
anonymous
  • anonymous
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: \(x = \pm \sqrt{\cosh^{2}y - y^{2}\coth^{2}y}\)
anonymous
  • anonymous
I'm not allowed to use that theorem. The idea behind this is to show it algebraically.
anonymous
  • anonymous
Can it be?? sinz =z --> sin z - z=0 iff z = 2kpi?
anonymous
  • anonymous
No, because then you would have \(\sin(2k\pi)- 2k\pi = 0 \implies 2k\pi = 0\)

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