Why is this proof wrong? $2^i = \frac{\sqrt{2}}{2}$

Question

If $i=-1^\frac{1}{2}$, then $2^i$ could be written as $2^{{-1}^{1/2}}$. By the multiplicative law of exponents, this should be equal to $2^{-1/2}$, which is $\frac{\sqrt2}{2}$.

This doesn't feel right. Where have I gone wrong?

Answer 1

Your argument seems to be $$2^i = 2^{\left(-1^{1/2}\right)} \color{red}{\stackrel{???}=} (2^{-1})^{1/2} = \left(\frac12\right)^{1/2} = \frac{\sqrt2}2.$$

But this is wrong, because even in straightforward cases $$a^{\left(b^c\right)} \ne \left(a^b\right)^c = a^{bc}.$$ Consider $$2^{\left(2^3\right)} \ne \left(2^2\right)^3 = 2^{2\cdot 3}$$ for example. The left side is $2^8=256$, but the right side is $2^6=64$.

There can be tricky issues with the $a^b$ notation when $a$ or $b$ is not a real number, but this is not one of the tricky cases. One can define $$2^x = \left(e^{\ln(2)}\right)^x = e^{x\ln(2)}$$ where $\ln$ is the natural logarithm function. The actual value of $\ln(2)$ is approximately $0.693$.

Then by Euler's formula, $$e^{ix} = \cos (x) + i\sin (x),$$ of which perhaps you have heard, we get:

$$\begin{align} 2^i & = \cos(\ln 2) + i\sin(\ln 2) \\ & \approx 0.769 + 0.639i.\\\end{align}$$

I suppose that this strange-looking result feels no better than the one you (correctly) rejected. Its actual meaning is a profound and extremely important matter. I'd like to recommend a book for you to pursue it further if you want to, but none is coming to mind. Tristan Needham's Visual Complex Analysis is superb, but may be too advanced for you at this stage.

Answer 2

To start with, it is problematic to try to use notation of the form $x^s$ and let $s$ be an object that is not an element of $\mathbb{Z}$. There is no consistent and unambiguous way to define this notation for $s\in\mathbb{C}$ without making the notation very unsatisfactory in general, so which "laws" of exponents hold in a given circumstances is very much dependent on context, and is purely a matter of semantic convenience, not a matter of actual mathematical truth.

That being said, one reasonable definition that people commonly use (though not used by mathematicians, because again, it is ultimately problematic) is that $z^{\zeta}=\exp(\zeta\log(z))$, where $\log(z)$ is understood to be short notation for $\ln(\left|z|\right)+\mathrm{atan}2\left(\mathrm{Im}(z),\mathrm{Re}(z)\right)i$ for $z\neq0$. In that context, $2^i=\exp(i\log(2))=\exp(i\ln(2))=\cos(\ln(2))+\sin(\ln(2))i$. Unlike the notation $2^i$, the notation $\exp(i\log(2))$ is not problematic, since $\exp$, $\ln$, and $\mathrm{atan}2$ are unambiguously, rigorously defined functions in complex analysis.