Question

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

Answer 1

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?

Answer 2

oh oh oh ok ok i think i got something!
a clever little trick.

lemme explain it and see if this is working.

Answer 3

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

Answer 4

\[\large\rm \color{orangered}{\sin(z)}=z\]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,\[\large\rm \color{orangered}{\frac{1}{2}\left(e^{i\left(\frac{\pi}{2}-z\right)}+e^{-i\left(\frac{\pi}{2}-z\right)}\right)}=z\]

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

Oo good call, I missed that in the denominator.

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

I guess I should've expected that :) lol

Answer 13

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.

Answer 14

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

Answer 15

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.

Answer 16

.

Answer 17

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:


\(x = \pm \sqrt{\cosh^{2}y - y^{2}\coth^{2}y}\)

Answer 18

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

Answer 19

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

Answer 20

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