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Where is the error in the following statement: $i^2=(i^2)^{\frac{4}{4}}=(i^4)^{\frac{2}{4}}=(1)^{\frac{1}{2}}=1$?

I feel the error is in the first equality, because $(i^2)^{\frac{4}{4}}$ is in fact $((i^2)^4)^{\frac{1}{4}}$ which is 1, but what is the mathematical reason so that raising a number to the $4/4$th power is not necessarily the number itself? Is it because the fourth root of a fourth power of a number $x$, may be either $x$ or $-x$, depending upon which is positive? I think I need a rigorous explanation.

Thanks for your time.

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Note, you aren't really using $i$, you could just as well write this as: $(-1)^{\frac{2}{2}} = ((-1)^2)^{\frac{1}{2}}$ –  Thomas Andrews Oct 12 '11 at 13:14
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possible duplicate of -1 is not 1, so where is the mistake? –  J. M. Nov 27 '11 at 2:22

3 Answers 3

up vote 2 down vote accepted

Ok, so rigorous explanation should be given in terms of complex analysis. But first consider even the simpler example: $-1 = (-1)^1=(-1)^{2/2} = \left((-1)^2\right)^{1/2} = 1^{1/2}=1$.

The trick is coming when you're take non-integer power of negative number or of complex one with a non-zero imaginary part. The reason is that you define $y=x^{1/m}$ as $$ y = \{z:z^m = x\} $$ and an equation $z^m=x$ has $m$ different roots. That's why when you're writing $a = a^{4/4}$ you implicitly take the forth root and so the answer is not uniquely defined. To be precise, you should think of it as $a\in a^{4/4}$ since ther right-hand side is not a unique element.

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$(1)^{\frac{1}{2}}$=1 or -1

So in general the square root gives two answers. + and -. Since essentially $x^2-1=0$ has two solutions x=1 and x=-1. So that is why your reasoning is flawed.

A very similar flawed argument like that is raising $(e^{i\theta})^i$ and assuming the usual power rules apply to it.

In general this is a big problem with complex analysis. Square root function, lnx, e^x and other can give many answers if you put complex numbers in them.

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Nitpicking: the exponential function $e^x$ is well-defined and single-valued over its entire domain; however, it's many-to-one (e.g., $e^{2\pi i} = e^0 = 1$) so its inverse $\ln x$ is (as you note) not single-valued. –  Steven Stadnicki Nov 27 '11 at 2:31

Noninteger exponents $x$ in power expressions $a^x$ are allowed only for bases $a>0$. Whenever a complex number not of this form appears as a basis the usual rules of handling powers are no longer valid.

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