Algebraic Manipulation of Euler's Identity Leads to a Strange Result

[amolv06]

I was playing around with Euler's identity the other day. I came across something that seems contradictory to everything else I know, but I can't really explain it.

I started with

$$e^{i\pi} = -1$$.

I rewrote this as

$$ln[-1] = i\pi$$

Multiplying by a constant, we have

$$kln[-1] = ki\pi$$

and using log properties I arrived at

$$ln[-1^{k}] = ki\pi$$

Now if I set k equal to any even number, I have

$$ln[1] = 0 = ki\pi$$

This seems to imply that $$i\pi$$ is 0, however it is not. Furthermore, any even value of k gives the same answer. Why is this?

 

[Diffy]

ln(-1) = (2/2)*ln(-1) = (1/2)*2*ln(-1) = (1/2)*ln(-1^2) = (1/2)*ln(1) = (1/2)*0 = 0

Therefore, ln(-1) = 0.

In order for properties of logs to work, you must be taking the log of a valid number. You can not take the log of a negative number, therefore, when you apply the properties of logs to something that is undefined, you get ludicrousness.

 

[dodo]

In the complex domain, log is multivalued, which is to say that it's not even a function in the first place.

You can read more about it http://en.wikipedia.org/wiki/Comple...9_as_the_inverse_of_the_exponential_function".

 

[amolv06]

Thanks!