π=0??(a question about e, i and π)

This contradictory result is due to the misunderstanding of the definition of natural logarithm for complex numbers.

Note that for natural logarithm of a complex number z, it is no longer a single-valued function, but a multi-valued function instead.

For any non-zero complex number z, the natural logarithm of z, denoted by Log(z), is the inverse of exponential function, i.e., e^[Log(z)] = z.
Note that by Euler formula, z = re^[i(2nπ + θ)], where r and θ are real numbers with r positive, and n integer.
Therefore e^[Log(z)] = re^[i(2nπ + θ)] = e^[ln(r) + i(2nπ + θ)], i.e., Log(z) = ln(r) + i(2nπ + θ), here ln denotes the natural logarithm for positive numbers.

This leads to the following definition for natural logarithm of complex numbers:

The natural logarithm of a non-zero complex number z, denoted by Log(z), is defined by
　　Log(z) = ln(|z|) + i[2nπ + arg(z)],
where |z| and arg(z) denote the modulus and the principal value of argument of z respectively.


Here comes back to your original problem.
From e^(2πi) = 1, taking natural logarithm on both sides,
ln[e^(2πi)] = 2nπi = ln(1), which no longer shows contradiction.

因為對數是定義在實數考慮的問題.
事實上歐拉公式是定義在複數去考慮問題：
e^(iθ)=cosθ+isinθ
當cosθ+isinθ=1時
θ=2nπ,(n為整數)
0是它的其中一個解,並不表示0=2π=4π=........

正如我們熟悉的一元二次方程
x^2-2x-3=0的解是3和-1
即3^2-2*3-3=(-1)^2-2*(-1)-3
那麼3=-1嗎??

希望能幫到你吧!

2007-03-11 19:31:09 補充：
因為對數的定義以a為底b的對數(其中a>0,且a≠1,b>0)在這樣的定義下它的對數必是實數而ln(-1)不合定義b>0.正如:x^4=16,它的根是±2或±2i但logx16=4,我們只能得出x=2道理一樣.

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