Where is the fallacy? $i=1$?

Question

Normally, when we have $\sqrt {-1}$, we let it be equal to $i$, an imaginary number. But what if we evaluate $\sqrt {-1}$ like this: $$\sqrt {-1}$$ $$(-1)^{\frac{1}{2}}$$ $$(-1)^{\frac{2}{4}}$$ $$((-1)^{2})^{\frac{1}{4}}$$ $$(1)^{\frac{1}{4}}$$ $$1$$ Okay, I know this is wrong. But I don't know why. I followed all of the rules of exponents... are there, perhaps, some special rules that need to be applies when the base is negative?

Answer 1

This isn't exclusive to complex numbers. Compare to this similar "proof":

$-1=(-1)^{2/2}=({(-1)^2})^{1/2}=1^{1/2}=1 .$

The issue is $({a^b})^c=a^{bc}$ only makes sense for fractional powers if you define the roots consistently, as explained in the comments. It is possible to define roots consistently for positive $a$ by always taking the positive root.

Answer 2

There is always a problem with defining $n^{1/m}$ when $n < 0.$ If we are asked for the cube root of $-8,$ we say fine, I know that, it must be $-2.$ Good so far.

If we attempt to define that as an actual exponentiation, we crash into the lack of a principal logarithm on the negative reals. If we could do it, we would have $$ (-8)^{1/3} = (-8)^{2/6} = \mbox{mess} $$

Answer 3

The problem is in the last line, you have that $(1)^{\frac{1}{4}}=1$, but here we have actually picked a specific $4^{\text{th}}$ root of $1$, and it's the wrong one.

In the complex plane we have that exponeniation is a multifunction. Note that for any $z \in \Bbb{C}$ we get $z=re^{i\theta}$ with $|z|=r$ and $\arg(z)=\theta$.
But we also have that $z=re^{i\theta+2\pi in}\;$ for $n \in \Bbb{Z}$. So we define the logarithm of $z$ as $\log(z) = \log(r) +i\theta +2\pi in$ which is a multifunction itself due to the fact that we can have any $n \in \Bbb{Z}$.
Now to deal with exponentiation, $a^b = \exp(b\log(a))$ we get that this too is a multifuction due to our definition of $\log$ in the complex plane.
So we can have to be careful when we make choices in the outcome of working out $a^b$ unlike how we can just do it in $\Bbb{R}$.

In your case $1^{\frac{1}{4}} = \exp(\frac{1}{4}\log(1))=\exp(\frac{1}{4}\times 2\pi i n)$ for $n \in \Bbb{Z}$.
This gives $1^{\frac{1}{4}}$ being one of $\{1, i, -1, -i\}$.

Answer 4

$1 ^ \frac{1}{4}= 1 $ has a problem. $1 ^ \frac{1}{4}$ has 4 distinct values, and they are $-1, 1, i, -i$. What you did is jumping from one solution to another, if you see your calculations as equations.

It is similar to this:

$$x^2-x+\frac{1}{4}=0$$ Obviously $x=-1/2$ but keep on arranging. $$x^2+\frac{1}{4}=x$$ $$x=x^2+\frac{1}{4}$$ Apply this new expression of $x$ to the starting equation. $$(x^2+\frac{1}{4})^2-(x^2+\frac{1}{4})+\frac{1}{4}=0$$ Let $y=x^2+\frac{1}{4}$ $$y^2-y+\frac{1}{4}=0$$ We know that $y=-1/2$. Then, $$y=x^2+\frac{1}{4}=-1/2$$ $$x^2+\frac{1}{4}=-1/2$$ $$x^2=-3/4$$ $$x=+/-\frac{i\sqrt3}{2}$$

Basically what you did was: $$x^2=-1$$ $$(x^2)^2=1$$ $$x^4=1$$ $$x=1$$

What is the general mistake here is that, taking bigger powers create uncertainty in the value, like creating more solutions in an equation; higher the degree, the more solutions. And taking roots collapse (some or all of) those values or solutions to a singular value. In my example, I started with 2 (identical) solutions as in 2nd degree polynomials, and then rose to a 4th degree equation (when $x^2$ and $x^{-2}$ exist together in an equation, just multiply both sides with $x^2$ and you will get a polynomial of degree 4). And finally I collapsed all those values into two values(which differ only by signs) that I did not start with, but that occured during moving from 2nd degree to 4th degree. Notice that the inital two identical solution and the last two solutions together are all correct solutions to the 4th degree equation that occured in the middle. The last step was just like you stating $x^4=1$ so $x=1$