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        riddles >> general problem-solving / chatting / whatever >> google of log(-1)
        
(Message started by: william wu on Dec 10^th, 2003, 10:26am)

Title: google of log(-1)
Post by william wu on Dec 10^th, 2003, 10:26am

When I type "log(-1)" into google, the google calculator spits out:

log(-1) = 1.36437635 i

I was wondering how google computes this. I think the answer should be [pi] i, using the formula:

ln(z) = ln(|z|) + i arg(z)

where z is a complex number. In this case, z = -1 = e^i[pi]. This is a vector with magnitude one and phase [pi]. Thus, we should get

ln(-1) = ln(1) + i([pi]) = [pi]i

I already asked the google people via e-mail about this a month ago, and I got some clerky response along the lines of we can't discuss how google computes results, but we forwarded this to our engineering team since this result may not be correct (and they haven't corresponded since). Maybe someone here can make sense out of 1.36437635 i ?

Title: Re: google of log(-1)
Post by TenaliRaman on Dec 10^th, 2003, 12:06pm

check the bases!

log(-1)
= log₁₀(-1)
= log₁₀(e^i*pi)
=i*pi*log₁₀(e)
=1.364376354*i

Title: Re: google of log(-1)
Post by towr on Dec 10^th, 2003, 12:46pm

yep..
To the general public log is the 10log
While around here it's generally considered the same as ln, the natural logarithm.. It can cause a lot of confusion if you're not carefull..

Title: Re: google of log(-1)
Post by TenaliRaman on Dec 10^th, 2003, 1:09pm

Actually even my scientific calculator follows the same (log as base 10 and ln as base e), so i always thought that it was a general notion.

However in mathematics log or ln is interchangeably used and assumed to be natural logarithm unless otherwise specified.(very similar to how x in sin(x) is assumed to be radians unless otherwise specified that it is in degrees).

Title: Re: google of log(-1)
Post by william wu on Dec 10^th, 2003, 2:32pm

Oh. Well that was a simple answer. Yeah, I should've known to check the bases *bonk* :P