Prove: $|xy|=|x|\cdot |y|$ Can you help me to prove this equality?
$$|xy|=|x|\cdot |y|$$
I don't know how to start proving the equality.
 A: Using the fact that $\lvert a \rvert = \sqrt{a^2}$ for all $a \in \mathbb{R}$, we obtain
$$
\begin{array}{rcl}
\lvert xy \rvert
  &=& \sqrt{ (xy)^2 } \\
  &=& \sqrt{ x^2 \cdot y^2 } \\
  &=& \sqrt{ x^2 } \cdot \sqrt{y^2} \\
  &=& \lvert x \rvert \cdot \lvert y \rvert
\end{array}
$$
for all $x,y \in \mathbb{R}$.
A: If $x$ and $y$ are both positive it is obviously true, and the case of one or both being zero is equally trivial.
If one of them is negative (say $y$), you have $$|xy|=x(-y)=x|y|=|x||y|.$$
If both are negative, you have $$|xy|=(-x)(-y)=|x||y|,$$
which exhausts the possibilities. 
A: There are three cases to consider depending whether each of $x$ and $y$ is negative, zero, or positive.  First, if either $x=0$ or $y=0$, we have  $$  |xy| = 0 = |x||y|  \text{.}  $$
Then, if both $x$ and $y$ have the same sign, $xy > 0$ and either  \begin{align}
x &< 0, y < 0, \text{ so }  & |xy| &= xy = (-x)(-y) = |x||y|  \text{ or}  \\
x &> 0, y > 0, \text{ so }  & |xy| &= xy = (x)(y) = |x||y|  \text{.}
\end{align}
Finally, if $x$ and $y$ have opposite signs, $xy < 0$ and either  \begin{align}
x &< 0, y > 0, \text{ so }  & |xy| &= -(xy) = (-x)(y) = |x||y|  \text{ or}  \\
x &> 0, y < 0, \text{ so }  & |xy| &= -(xy) = (x)(-y) = |x||y|  \text{.}
\end{align}
A: You are assuming $x$, $y$ are real. To prove for all cases, you must say $x=a+bi$, $y=c+di$. Then you have
$$|xy|=|(ac-bd)+(ad+bc)i|=$$
$$=\sqrt{a^2c^2+b^2d^2-2abcd+a^2d^2+b^2c^2+2abcd}$$
$$=\sqrt{a^2c^2+a^2c^2+b^2c^2+b^2d^2}$$
$$=\sqrt{a^2+b^2} \cdot \sqrt{c^2+d^2}=|x||y|$$
