Am I right in saying that the absolute value symbols act like a function such that if $x$, for example, is $x<0$ then $x=-x$
In other words $x$ will be positive regardless of what value you give to $x$, right?
In which case, isn't $|z|$ always positive? Considering $|z|=\sqrt{a^2+b^2}$, where $z=a+bi$.
Yes, the absolute value of a number is always non-negative (Zero if the input is zero, positive otherwise).
This is the point of the absolute value function; it is in some sense a measure of the size of the number. To be more precise, the absolute value of a number is the distance from the number to the origin.
Am I right in saying that the absolute value symbols act like a function such that if $x$, for example, is $x<0$ then $x=-x$
Yes, if you would add bars: $\lvert x \rvert = -x$ for $x < 0$.
The definition is $$ \lvert x \rvert = \left\{ \begin{array}{cccl} x & \mbox{ for } & x > 0 & \Rightarrow \mbox{positive sign} \\ 0 & \mbox{ for } & x = 0 & \Rightarrow \mbox{no sign} \\ -x & \mbox{ for } & x < 0 & \Rightarrow \mbox{positive sign} \\ \end{array} \right. $$ Therefore $\lvert x \rvert$ is always positive or zero, thus non-negative.
In other words $x$ will be positive regardless of what value you give to $x$, right?
No, $\lvert 0 \rvert = 0$, which is not positive and not negative.
In which case, isn't $|z|$ always positive? Considering $|z|=\sqrt{a^2+b^2}$, where $z=a+bi$.
No, if $r = 0$, thus $z=0$, then $\lVert z \rVert = 0$, which is not positive.
Now we are talking about the complex absolute value, which can be viewed as 2-norm on $\mathbb{R}^2$: $$ \lVert z \rVert^2 = z \bar{z} = (\mbox{Re } z)^2+ (\mbox{Im } z)^2 $$ Among the properties it shares with the real version are: $$ \lVert z \rVert \ge 0 \\ \lVert z \rVert = 0 \iff z = 0 $$
