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# Logarithms of negative numbers

_AVG_
They've fascinated me for quite a while.

I've a couple of queries :

First, do they exist?

Second, if they do exist, is this equation correct : ln(-1)=pi*i

Third, is there any way to compute logarithms of complex numbers?

Bikerman
 _AVG_ wrote: They've fascinated me for quite a while. I've a couple of queries : First, do they exist?
Yep - they give complex number solutions.
 Quote: Second, if they do exist, is this equation correct : ln(-1)=pi*i
Correct.
 Quote: Third, is there any way to compute logarithms of complex numbers? Fourth, where could I find more information on them?

http://en.wikipedia.org/wiki/Complex_logarithm
Voodoocat
The reason that the log of a negative number is imaginary not real is due to the definition of logs:

log base 2 of 8 = 3 because 2^3 = 8.

The problem with negative logs is that a negative exponent does not mean that the number is negative, it means that the number is in the wrong place: 2^3 = 8, but 2^-3= 1/8

Hence, the log of a negative number is a complex number
_AVG_
 Voodoocat wrote: The reason that the log of a negative number is imaginary not real is due to the definition of logs: log base 2 of 8 = 3 because 2^3 = 8. The problem with negative logs is that a negative exponent does not mean that the number is negative, it means that the number is in the wrong place: 2^3 = 8, but 2^-3= 1/8 Hence, the log of a negative number is a complex number

I'm sorry ... I don't quite grasp your explanation. Could you explain in more detail?
Don't negative exponents also yield positive results? Only if the base is negative will the result be negative right no matter what the exponent is?
pscompanies
_AVG_ wrote:
 Voodoocat wrote: The reason that the log of a negative number is imaginary not real is due to the definition of logs: log base 2 of 8 = 3 because 2^3 = 8. The problem with negative logs is that a negative exponent does not mean that the number is negative, it means that the number is in the wrong place: 2^3 = 8, but 2^-3= 1/8 Hence, the log of a negative number is a complex number

I'm sorry ... I don't quite grasp your explanation. Could you explain in more detail?
Don't negative exponents also yield positive results? Only if the base is negative will the result be negative right no matter what the exponent is?

I think what Voodoocat is trying to say is that if log base x of y = a, then x^a = y.

Now, you can see the problem that if y < 0 no a exists such that a number x raised to the power a can equal a negative number. For example,
(3)^-2 = 1/(3)^2 > 0 and (-3)^2 = 9 > 0 and (-3)^-2 = 1/(-3)^2 = 1/9 > 0.

This is why you have to use complex numbers.
infinisa
Hello All

Quite frankly, I think Bikerman said all that needs to be said on this question.

I suggest that before anyone posts anything else, they should read his post, and only then post something if they have some difficulty understanding it.
iman
If you try to graph y = x^n where x is negative and n is an integer, you'll get a crazy graph.

If you include all real n, I don't know how that looks like in the real plane.
metalfreek
pscompanies wrote:
_AVG_ wrote:
 Voodoocat wrote: The reason that the log of a negative number is imaginary not real is due to the definition of logs: log base 2 of 8 = 3 because 2^3 = 8. The problem with negative logs is that a negative exponent does not mean that the number is negative, it means that the number is in the wrong place: 2^3 = 8, but 2^-3= 1/8 Hence, the log of a negative number is a complex number

I'm sorry ... I don't quite grasp your explanation. Could you explain in more detail?
Don't negative exponents also yield positive results? Only if the base is negative will the result be negative right no matter what the exponent is?

I think what Voodoocat is trying to say is that if log base x of y = a, then x^a = y.

Now, you can see the problem that if y < 0 no a exists such that a number x raised to the power a can equal a negative number. For example,
(3)^-2 = 1/(3)^2 > 0 and (-3)^2 = 9 > 0 and (-3)^-2 = 1/(-3)^2 = 1/9 > 0.

This is why you have to use complex numbers.

Great explanation.
panpanman
The imaginary plane has a lot of cool stuff in it, especially...

e^(i*pi)=-1

http://xkcd.com/179/

kelseymh
 _AVG_ wrote: Third, is there any way to compute logarithms of complex numbers?

This was the only part Bikerman didn't directly address, but you can derive it fairly easily. Use exponential notation, i.e., write <b>a+ib</b> as

<b>AcosT + iAsinT = A exp(iT)</b>,

where

<b>tanT = b/a</b>

computed over the full domain [0,2pi) taking account of the signs of both <b>a</b> and <b>b</b>, and

<b>A = a/cosT = b/sinT</b>

(equal by construction, dealing with the limiting cases in the obvious way). Then the solution is trivial:

<b>ln(a+ib) = ln[A exp(iT)] = ln A + iT</b>

Here <b>A</b> is positive definite by construction, since the computation of <b>T</b> ensures that <b>cosT</b> has the same sign as <b>a</b>.