Cos(x)=-2 DEFINED?

  • Thread starter GreenPrint
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  • #26
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-i ln ( -2 + sqrt(3) )
raised to -1
i ln( (-2 + sqrt(3))^-1 )
simplified
i ln( 1/(-2 + sqrt(3)) )
Multiplied by conjerget
i ln( (-2 - sqrt(3))/( (-2 + sqrt(3))(-2 - sqrt(3)) ) )
simplify
i ln( (-2 - sqrt(3))/(4 + 2sqrt(3) - 2sqrt(3) - 3)
i ln( (-2 - sqrt(3))/(4 - 3)
i ln( (-2 - sqrt(3) )/1)
?
 
  • #27
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oh wow let me see here let me reread
 
  • #28
tiny-tim
Science Advisor
Homework Helper
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Hi GreenPrint! :smile:

(what happened to that π and √ i gave you? :confused:)

Mentallic was right … I got a minus in the wrong place. :redface:

I've edited my last post to correct it.
 
  • #29
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Don't SIMPLIFY anything! Just follow the formulas. Also, why don't you try clicking on some of my formulas to see what the latex code looks like. It helps a lot in terms of readability of your equations.
 
  • #30
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wow ok the thing confused there is that the value that I solved for was actually a side length and you are solving for an angle now??? That's kind of werid but I made sense fo the formulas after realizing that... my only question is why didn't you use +/- and just used +
 
  • #31
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then I just plug in what I got for my answers into the formula? where my answer is the z? Wow well thanks for the help hmmm that's interesting let me see what I get...
 
  • #32
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No, your answer is the [itex]w[/itex] and [itex]z = -2[/itex].
 
  • #33
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what are you using for that slanted p value? let me look again hold up
 
  • #34
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my only question is why didn't you use +/- and just used +

The complex square root is a double - valued function:

[tex]
z = w^{2}, \; z = \rho \, e^{i \, \theta}, \; w = R e^{i \Theta}
[/tex]

[tex]
\rho = R^{2} \Rightarrow R = \sqrt{\rho}
[/tex]

[tex]
\theta + 2 \, n \, \pi = 2 \, \Theta \Rightarrow \Theta = \frac{\theta}{2} + n \, \pi, \; n \in \mathbb{Z}
[/tex]

But, [itex]\Theta[/itex] is restricted to a period of [itex]2 \, \pi[/itex] so:

[tex]
\sqrt{z} = \left\{\begin{array}{l}
\sqrt{\rho} \, e^{i \, \frac{\theta}{2}} \\

\sqrt{\rho} \, e^{i \, (\frac{\theta}{2} + \pi)}
\end{array}\right.
[/tex]
 
Last edited:
  • #35
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ok well thanks I'll post post back if i get stuck again i need to do some more stuyding of this stuff :)
 
  • #36
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But, your initial hunch was correct. The equation in the topic of the thread does have (infinitely many) solution(s) in the set of complex numbers [itex]\mathbb{C}[/itex].
 
  • #37
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So my final answers should be
2 pi n - i(-2 + sqrt(3))
and
2 pi n -i(-2 - sqrt(3))

where n is the set of integers?
 
  • #38
2,981
5
So my final answers should be
2 pi n - i(-2 + sqrt(3))
and
2 pi n -i(-2 - sqrt(3))

where n is the set of integers?

Where is the [itex]\ln[/itex] function and what is the absolute value and argument of:

[tex]
-2 + \sqrt{3}
[/tex]

and

[tex]
-2 - \sqrt{3}
[/tex]
 
  • #39
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sorry I forgot the ln
so is this correct
2pi n - i ln(2 + sqrt(3))

and i take the absolute value of the argument in the natural log leaving me with only positives? where n is the set of integers
 
  • #40
2,981
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argument of a complex number [itex]z = \rho \, e^{i \, \varphi}[/itex] is the variable [itex]0 \le \varphi < 2 \pi[/itex]. What is it for the two inputs of the complex logarithm?
 
  • #41
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2 and sqrt(3)... right?
 
  • #42
2,981
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no. you need to look up trigonometric form of a complex number.
 
  • #43
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ok and what exactly am I looking for
 
  • #44
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z = |z|[cos (t) + i sin (t) ]
 
  • #45
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right it's the cosine and sine of the angle?
 
  • #46
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i(theta) ?
 
  • #47
hunt_mat
Homework Helper
1,745
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If cos(z)=-2, then just use:
[tex]
\cos z=\cos (x+yi)=\cos x\cos (yi)-sin x\sin (yi)=\cos x\cosh y-i\sin x\sinh y=-2
[/tex]
Compare real and imaginary parts to obtain:
[tex]
\cos x\cosh y=-2\quad\sin x\sinh y=0
[/tex]
Clearly x must be a multiple of pi and that you can work you the solution for what y has to be.
 

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