What is the explanation for the complex logarithms function title?

[GreenPrint]

Homework Statement

When was trying to solve this problem
cos(x)=-2 I got the following answers
i ln(2 +/- sqrt(3)) + 2 pi n
n set of integers positive and negative
now I was told by a couple of people on here to add 2 pi n to my solution because before it didn't have it. At the time I didn't understand why but now I do. All I did was simply think about it as an angle and hence adding multiples of 2 pi n give valid solutions but what I don't understand is why does
http://www.wolframalpha.com/input/?i=(i^((2x)/pi)+i^((-2x)/pi))/(2)=-2
give me the solutions of
x = 2 pi n + pi - i log(2 +/- sqrt(3) ), n set of positive and negative integers ?
Now please understand I get that there are an infinite number of solutions. I completely understand this. This is the reason why when we define a function as a mathematical formula or representation that only has one output value for each input value of it's domain is false, correct? The complex logarithm is a completley valid function. So I understand there are an infinite amount of solutions I just don't understand were that other pi came from or the -i. I was woundering if somebody cuold please explain this to me. Thank You!

Homework Equations

The Attempt at a Solution

 

[GreenPrint]

anybody please help?

 

[Char. Limit]

You got your answer wrong. The inverse of cos(y)=x is

$$y=\frac{log\left(x\pm\sqrt{x^2-1}\right)}{i}$$

Can you spot your mistakes? There are two.

 

[GreenPrint]

Iw asn't dong the inverse of cosine though sorry I forgot to mention it I was doing this

cos(x)=-2

so you don't have that domain restriction correct?

 

[GreenPrint]

and I have no idea were that equation came from... also it's not normal to divide by i right

 

[Char. Limit]

and I have no idea were that equation came from... also it's not normal to divide by i right

I actually messed it up and was fixing it when you posted.

The domain restriction doesn't apply here, and dividing by i is just multiplying by -i... it's necessary.

 

[GreenPrint]

ok well unforunatley i don't know were the mistakes are =(...

 

[Char. Limit]

ok well unforunatley i don't know were the mistakes are =(...

First, you multiplied by i instead of -i.

Second, the 2 in your answer should be a -2.

After a bit of playing with logarithmic properties, I managed to turn one of the answers into W-A's answer.

 

[GreenPrint]

Tahnk youu

 

[GreenPrint]

still need help

Ok well wolfram's answer was this
x = 2 pi n + pi -i log(2-sqrt(3)), n element Z
and
x = 2 pi n + pi -i log(2+sqrt(3)), n element Z

now when i solved it I got this for my answers
x = iln(2+3^.5) + 2 pi n
and
iln(2-3^.5) + 2 pi n
n set of integers positive and negative

the two is not negative and I still don't know where that other pi came from in wolfram's answer can you please help

 

[Char. Limit]

still need help

Ok well wolfram's answer was this
x = 2 pi n + pi -i log(2-sqrt(3)), n element Z
and
x = 2 pi n + pi -i log(2+sqrt(3)), n element Z

now when i solved it I got this for my answers
x = iln(2+3^.5) + 2 pi n
and
iln(2-3^.5) + 2 pi n
n set of integers positive and negative

the two is not negative and I still don't know where that other pi came from in wolfram's answer can you please help

Well, let's start with my answer, given by my inverse formula for cos(x)...

$$i\left(-log\left(-2\pm\sqrt{3}\right)\right)$$

Now if you factor the -1 out of the inside and then use the rule $alog(b)=log(b^a)$, you get this:

$$ilog\left(\frac{-1}{2\pm\sqrt{3}}\right)$$

Which after using the rule $log\left(\frac{a}{b}\right)=log(a)-log(b)$ becomes...

$$ilog(-1)-ilog(2\pm\sqrt{3})$$

From there it's just recognizing that $ilog(-1)=-\pi=\pi+2\pi n$ for n=-1, an integer.