Simplify Natural Logarithms: ln(i^2)^2 Explained | Sweet Bro"

[FUNKER]

what is the result or simplification of

( ln(i^2) )^2

sweet bro

 

[Janitor]

Easy as pi

First, i^2=-1.

Then note that exp(i pi) = -1.

So ln(-1) = ln[exp(i pi)] = i pi.

Square it: (i pi)^2 = - pi^2.

EDIT: I should caution that this is a solution. I make no claim that it is the only solution.

 

[chroot]

$-\pi^2$ is definitely a solution. I'm not sure how there could be more solutions...?

- Warren

 

[cookiemonster]

He wasn't even referring to solutions, was he? He just wanted a simplification of an expression, no?

cookiemonster

 

[chroot]

Well, heh, right, not solutions... I meant that $-\pi^2$ is the simplest possible form for it. I see what Janitor was saying now -- that there may be better simplifications. If so, I don't see any.

- Warren

 

[Janitor]

Cookiemonster: he did ask, "What is the result..."

I'm not sure how there could be more solutions...?

Chroot, here's the sort of thing I am worried about:

Note that exp(3 i pi) = -1.

So ln(-1) = ln[exp(3 i pi)] = 3 i pi.

Square it: (3 i pi)^2 = - 9 pi^2.

 

[matt grime]

the principal branch is usually what we mean when we ask for log of a complex number, which is the same as the principal value of arg, ie Arg

 

[FUNKER]

thanks for all ur help dudes, and yea I believe this is a solution as well as a simplification.

 

[Hurkyl]

When I took complex analysis, we used ln for the multivalued version and Ln for the principal branch.