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Solving a complex trigonometric equation

  • Thread starter frozenguy
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Hi, thank you for looking.
1. Homework Statement
Solve the following complex equation and carefully plot at least 3 solutions for each. State ALL solutions to each! State exact answers!
Sin(z)=√(3)

3. The Attempt at a Solution
Here is my work.

I was a little confused how to tell what function from each case (A or B) I should set equal to solve for the variable. I found sin(x) ≠ √(3) so went with cosh(y)=√(3)

Since I had an answer for y I went with cos(x) = 0 for the next instead of sinh(y) = 0.

Is this a reasonable process?


MATHW2prob3-1.jpg
 
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LCKurtz

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You have sin(x)cosh(y) = [itex]\sqrt 3[/itex]. While it is true enough that
[itex]\sin(x) \ne \sqrt 3[/itex] why do you conclude that cosh(y) = [itex]\sqrt 3[/itex]? You need to re-think that step and keep in mind that this is a simultaneous system of equations you are solving.
 
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You have sin(x)cosh(y) = [itex]\sqrt 3[/itex]. While it is true enough that
[itex]\sin(x) \ne \sqrt 3[/itex] why do you conclude that cosh(y) = [itex]\sqrt 3[/itex]? You need to re-think that step and keep in mind that this is a simultaneous system of equations you are solving.
Well I thought it was an odd step as well but it was similar to an example the professor did on the board..
I guess I don't know how else to approach the problem from there. You can't solve for/sub x or y between the imaginary and real parts can you?
 

LCKurtz

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Remember the x and y values you get must satisfy both equations. Since you can figure out what x and y can satisfy your second equation, that limits what you can try in the first equation.
 
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Ok, well sinh(y)=0 therefore y=0.

But if y=0, then cosh(y)=1 and would require sin(x)=√(3)
 

LCKurtz

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Ok, well sinh(y)=0 therefore y=0.

But if y=0, then cosh(y)=1 and would require sin(x)=√(3)
Right, so that doesn't work. There are two things that make equation B work and you have just ruled out one for equation A. What about the other one?
 
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Sinh(y)=0 only at y=0.

So cos(x)=0 at [itex]\frac{k\pi}{2}[/itex] with k=1,3,5,...
But then its the same as the original assumption I made in A.
Because if I sub [itex]\frac{\pi}{2}[/itex] for x in sin(x) I get 1. or -1 for k=3,7,11
 
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LCKurtz

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Sinh(y)=0 only at y=0.

So cos(x)=0 at [itex]\frac{k\pi}{2}[/itex] with k=1,3,5,...
But then its the same as the original assumption I made in A.
Because if I sub [itex]\frac{\pi}{2}[/itex] for x in sin(x) I get 1. or -1 for k=3,7,11
Yes. So now you are getting close to understanding and giving me the answer to my original question in first post. You have shown that B is satisfied if either sinh(y) = 0 or cos(x) = 0. So those are the only candidates for A. Then you showed y = 0 doesn't lead anywhere. So now you have the only possible values for x, and hence for sin(x). Since sin(x) can be ±1, that both explains why you set that equation [itex]=\sqrt 3[/itex] in the first place and corrects it to using [itex]\pm\sqrt 3[/itex].

Also, another notation thing you might like is to write the values for x like this:

[tex]x = \frac \pi 2 + k\pi,\ k \hbox{ integer.}[/tex]
to avoid the every other integer thing.
 
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Yes. So now you are getting close to understanding and giving me the answer to my original question in first post. You have shown that B is satisfied if either sinh(y) = 0 or cos(x) = 0. So those are the only candidates for A. Then you showed y = 0 doesn't lead anywhere. So now you have the only possible values for x, and hence for sin(x). Since sin(x) can be ±1, that both explains why you set that equation [itex]=\sqrt 3[/itex] in the first place and corrects it to using [itex]\pm\sqrt 3[/itex].

Also, another notation thing you might like is to write the values for x like this:

[tex]x = \frac \pi 2 + k\pi,\ k \hbox{ integer.}[/tex]
to avoid the every other integer thing.
Ahh thank you.. It's clicking now.

So when I go about graphing this,
It will be a set of points spread apart by [itex]\frac{\pi}{2}+k\pi[/itex] w/ k= int. at y=ln[2√(3)]


Thank you again for your help.
 
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