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

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The expression (ln(i^2))^2 simplifies to -π^2, as i^2 equals -1 and ln(-1) can be expressed using the exponential function. The discussion highlights that ln(-1) can yield multiple values, such as iπ and 3iπ, leading to different squared results. However, -π^2 is identified as the simplest form of the expression. Participants clarify the distinction between solutions and simplifications, emphasizing that while there may be multiple logarithmic values, -π^2 remains a valid simplification. The conversation underscores the importance of understanding the principal branch of the logarithm in complex analysis.
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what is the result or simplification of

( ln(i^2) )^2

sweet bro
:confused:
 
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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.
 
Last edited:
-\pi^2 is definitely a solution. I'm not sure how there could be more solutions...?

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

cookiemonster
 
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
 
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.
 
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
 
thanks for all ur help dudes, and yea I believe this is a solution as well as a simplification.
 
When I took complex analysis, we used ln for the multivalued version and Ln for the principal branch.
 

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