Equation

[Spin]

how to solve this: sin(z)= i

where z =x+iy.

 

(or if we have sin(z) = ni where n belongs to N(natural number) in general.

It's a question of a quiz and I don't know what kind of equation it is)

 

Thanks in advance

[Tom Mattson]

how to solve this: sin(z)= i

where z =x+iy.

 

Compare equations (2) and (3) in the entry on Sine at MathWorld and look at what happens when you make the substitution [imath]z \longrightarrow iz[/imath] in (3). You basically get (2)' date=' up to a constant. That means that can express [imath']sin(z)[/imath] in terms of [imath]sinh(iz)[/imath].

 

How does that help?

 

Look at equation (1) in the entry on Inverse Hypberbolic Sine at MathWorld. Once you make the substitution from the first paragraph you should be able to solve the equation with the result cited in this paragraph.

[s pepperchin]

Compare equations (2) and (3) in the entry on Sine at MathWorld and look at what happens when you make the substitution [imath]z \longrightarrow iz[/imath] in (3). You basically get (2)' date=' up to a constant. That means that can express [imath']sin(z)[/imath] in terms of [imath]sinh(iz)[/imath].

 

How does that help?

 

Look at equation (1) in the entry on Inverse Hypberbolic Sine at MathWorld. Once you make the substitution from the first paragraph you should be able to solve the equation with the result cited in this paragraph.

 

What is the point of that substitution? Where is that taking us. I see that if you substitute iz for z we get the sin(iz)=sinh(z)/i, but I don't see how this gets us to sin(z)=i, and there is no mention of the z=x+iy.

[Tom Mattson]

What is the point of that substitution? Where is that taking us. I see that if you substitute iz for z we get the sin(iz)=sinh(z)/i' date=' but I don't see how this gets us to sin(z)=i,

[/quote']

 

It doesn't get us there. But then again that's not the substitution I advised. I said to let [imath]z \rightarrow iz[/imath] in equation (3), not equation (2). If you do that then you get the following.

 

[math]\sinh(iz)=\frac{e^{iz}-e^{iz}}{2}=i\sin(z)[/math]

 

Solve that for [imath]\sin(z)[/imath], substitute into the original equation, and then solve for [imath]\sinh(iz)[/imath]. You can then use the expression for the inverse hyperbolic sine from the second link I gave to get an explicit expression for [imath]iz[/imath], and then for [imath]z[/imath]. That solves the problem.

 

and there is no mention of the z=x+iy.

 

The definition of z as a complex variable is implicit in the original equation. The relation [imath]f(x)=\sin(x)[/imath] is a map from [imath]\mathbb{R}[/imath] to [math][-1,1][/math] when [imath]x[/imath] is a real variable. The only way to get [imath]\sin(z)=i[/imath] is if you have [imath]z=x+iy[/imath]. So yes, it is mentioned in that the entire problem is predicated on it.

[s pepperchin]

I'm looking at these websites and apparently I am missing a step, but I would appreciate it if you could show me the proof you used to solve this problem.

[Tom Mattson]

OK, first of all there was a typo in my last post. I forgot to include the [imath]\frac{1}{2}[/imath] in the definition of [imath]\sinh(z)[/imath]. It's fixed now.

 

Here are all the steps I used to solve the problem.

 

1.) Sub [imath]iz[/imath] for [imath]z[/imath] in [imath]\sinh(z)[/imath].

2.) Express [imath]\sin(z)[/imath] in terms of [imath]sinh(iz)[/imath] and substitute the expression in the equation.

3.) Solve the equation for [imath]\sinh(iz)[/imath].

4.) Take the inverse hyperbolic sine of both sides.

5.) Solve for [imath]z[/imath].

 

Now be advised that my first post to this thread was just the first approach that popped into my head. There is a much more direct way to solve this problem: Just take the inverse sine of both sides (use eq 1 in that link).

 

Why did I first think of the most obtuse way to do it? Beats me, but you get the same answer either way: [imath]z=-i\ln(-1+\sqrt{2})[/imath].