Confused with imaginary numbers In 9th grade I had an argument with my teacher that 
${i}^{3}=i$
where $i=\sqrt{-1}$
But my teacher insisted (as is the accepted case) that:
${i}^{3}=-i$
My Solution:
${i}^3=(\sqrt{-1})^3$
${i}^3=\sqrt{(-1)^3}$
${i}^3=\sqrt{-1\times-1\times-1}$
${i}^3=\sqrt{-1}$
${i}^3=i$
Generally accepted solution:
${i}^3=(\sqrt{-1})^3$
${i}^3=\sqrt{-1}\times\sqrt{-1}\times\sqrt{-1}$
${i}^3=-\sqrt{-1}$
${i}^3=-i$
What is so wrong with my approach? Is it not logical?
I am using the positive square root. There seems to be something about the order in which the power should be raised? There must be a logical reason, and I need help understanding it.
 A: Here's a better proof than the given "generally accepted proof":
$i^3 = i^2 \cdot i = -1 \cdot i = -i$
A: I think it is always healthier to avoid square roots. If we had equality $i^3=i$, then
$$
-1=i\times i=i\times i^3=i^2\times i^2=(-1)(-1)=1. 
$$
Or, you could simply check: $i^3=i^2\times i=(-1)i=-i$.
A: In a nutshell the square root function is not single valued.  So you cannot always say $\sqrt{ab}=\sqrt a\cdot \sqrt b$.  Otherwise you could also prove $1=-1$ as follows: $1=\sqrt1=\sqrt{-1\cdot-1}=\sqrt{-1}\sqrt{-1}=(\sqrt{-1})^2=-1$.
A: The error is in saying $i=\sqrt{-1}$. The correct statement is $i^2=-1$ and we're only allowed to use this property.
Thus
$$
i^3=i^2\cdot i=(-1)\cdot i=-i
$$
Never substitute $\sqrt{-1}$ for $i$: it just leads to errors, because the standard identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ just holds for real and non negative numbers.
A: The number $\sqrt{-1}$ is a two valued function with values $\{i,-i\}$ and hence the statement
$$
\sqrt{-1}=i
$$
is not true.
If you define the square function on the principal branch, then you can write $\sqrt{-1}=i$ but you can't pass the power into the square.
