Why is $i^3$ (the complex number "$i$") equal to $-i$ instead of $i$? [duplicate]

Since $i^2=-1$ by definition, $i^3=i^2\cdot i=-i$.

$\sqrt{a}\sqrt{b}=\sqrt{ab}$ is only guaranteed for positive real $a$ and $b$.

We cannot say that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ for negative $a$ and $b$. If this were true, then $1=\sqrt{1}=\sqrt{\left(-1\right)\cdot\left(-1\right)} = \sqrt{-1}\sqrt{-1}=i\cdot i=-1$. Since this is false, we have to say that $\sqrt{a}\sqrt{b}\neq\sqrt{ab}$ in general when we extend it to accept negative numbers.

I'm sure everyone has answered the question appropriately. But here's my 2 cents:

From the Argand plane perspective, multiplying a complex number by $i$ is equivalent to rotating it about a circle (with radius = modulus of complex number) counterclockwise by 90 degrees. So ask yourself where you end up when you take $i$ and multiply it with $i$ twice.

$\sqrt{ab} = \sqrt{a} \sqrt{b}$ is correct in the following sense: In an arbitrary field (here it is the field of complex numbers) the root $\sqrt{a}$ is an element in some field extension such that $\sqrt{a}^2=a$. It is not uniquely determined, for if $b$ is a root, then also $-b$ is a root (and these only coincide when $a=0$ or the characteristic is $2$). Now the correct statement is:

• If $\sqrt{a}$ and $\sqrt{b}$ are roots of $a$ resp. $b$, then $\sqrt{a}\sqrt{b}$ is a root of $ab$.

If we define $\sqrt{a}$ to be the set of all roots of $a$, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$ even holds verbatim. For example, $\sqrt{-1}=\{\pm i\}$ then, and $\sqrt{(-1) (-1)} = \sqrt{-1} \sqrt{-1}$ holds since both sides equal $\{\pm 1\}$.

If $a \in \mathbb{R}^+$, one usually denotes by $\sqrt{a}$ the unique root of $a$ in $\mathbb{R}^+$, but this definition doesn't work properly for complex numbers or other fields.