Solve Cos(x)=-2: Step-by-Step Guide

[GreenPrint]

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...

 

[Dickfore]

No, your answer is the w and z = -2.

 

[GreenPrint]

what are you using for that slanted p value? let me look again hold up

 

[Dickfore]

my only question is why didn't you use +/- and just used +

The complex square root is a double - valued function:

<br /> z = w^{2}, \; z = \rho \, e^{i \, \theta}, \; w = R e^{i \Theta}<br />

<br /> \rho = R^{2} \Rightarrow R = \sqrt{\rho}<br />

<br /> \theta + 2 \, n \, \pi = 2 \, \Theta \Rightarrow \Theta = \frac{\theta}{2} + n \, \pi, \; n \in \mathbb{Z}<br />

But, \Theta is restricted to a period of 2 \, \pi so:

<br /> \sqrt{z} = \left\{\begin{array}{l}<br /> \sqrt{\rho} \, e^{i \, \frac{\theta}{2}} \\<br /> <br /> \sqrt{\rho} \, e^{i \, (\frac{\theta}{2} + \pi)}<br /> \end{array}\right.<br />

 

[GreenPrint]

ok well thanks I'll post post back if i get stuck again i need to do some more stuyding of this stuff :)

 

[Dickfore]

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 \mathbb{C}.

 

[GreenPrint]

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?

 

[Dickfore]

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 \ln function and what is the absolute value and argument of:

<br /> -2 + \sqrt{3}<br />

and

<br /> -2 - \sqrt{3}<br />

 

[GreenPrint]

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

 

[Dickfore]

argument of a complex number z = \rho \, e^{i \, \varphi} is the variable 0 \le \varphi &lt; 2 \pi. What is it for the two inputs of the complex logarithm?

 

[GreenPrint]

2 and sqrt(3)... right?

 

[Dickfore]

no. you need to look up trigonometric form of a complex number.

 

[GreenPrint]

ok and what exactly am I looking for

 

[GreenPrint]

z = |z|[cos (t) + i sin (t) ]

 

[GreenPrint]

right it's the cosine and sine of the angle?

 

[GreenPrint]

i(theta) ?

 

[hunt_mat]

If cos(z)=-2, then just use:
<br /> \cos z=\cos (x+yi)=\cos x\cos (yi)-sin x\sin (yi)=\cos x\cosh y-i\sin x\sinh y=-2<br />
Compare real and imaginary parts to obtain:
<br /> \cos x\cosh y=-2\quad\sin x\sinh y=0<br />
Clearly x must be a multiple of pi and that you can work you the solution for what y has to be.