# Cos(x)=-2 DEFINED?

-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)
?

oh wow let me see here let me reread

tiny-tim
Science Advisor
Homework Helper
Hi GreenPrint!

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

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

I've edited my last post to correct it.

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.

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 +

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...

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

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

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

The complex square root is a double - valued function:

$$z = w^{2}, \; z = \rho \, e^{i \, \theta}, \; w = R e^{i \Theta}$$

$$\rho = R^{2} \Rightarrow R = \sqrt{\rho}$$

$$\theta + 2 \, n \, \pi = 2 \, \Theta \Rightarrow \Theta = \frac{\theta}{2} + n \, \pi, \; n \in \mathbb{Z}$$

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

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

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

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 $\mathbb{C}$.

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?

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:

$$-2 + \sqrt{3}$$

and

$$-2 - \sqrt{3}$$

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

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

2 and sqrt(3)... right?

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

ok and what exactly am I looking for

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

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

i(theta) ?

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