## 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

1. 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?

2. zepdrix

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

$\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

$\large\rm \color{orangered}{\sin(z)}=z$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

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

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

7. zepdrix

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

8. zepdrix

Oo good call, I missed that in the denominator.

9. zepdrix

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

10. 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}$ $\large\rm e^{iz}-e^{-iz}=2iz\qquad\qquad (2)$

11. 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.

12. zepdrix

I guess I should've expected that :) lol

13. 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.

14. 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

15. 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.

16. IrishBoy123

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17. 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: $$x = \pm \sqrt{\cosh^{2}y - y^{2}\coth^{2}y}$$

18. anonymous

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

19. anonymous

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

20. anonymous

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