If $f = u + iv$ is a complex function, is $|f| = (u^2 + v^2)^{1/2}$? If $f = u + iv$ is a complex function, is $|f| = (u^2 + v^2)^{1/2}$? Where, $|f|$ is the modulus or absolute of $f$.
I thought this should be correct because for any complex number $z = a + ib$, $|z| = (a^2 + b^2)^{1/2}$. At any point $w = x + iy$, $f(w) = u(x,y) + iv(x,y)$ is just a complex number and so the modulus should defined the same way.
 A: I misunderstood your question the first time.
You are asking: If $f=u+iv$, does the formula $|f|=(u^2 + v^2)^{1/2}$ hold?
First, your question is not well posed, because you have not said what $u$ and $v$ are. In fact in your first paragraph, you write $f=u+iv$, implying that $f$ is a function of two variables $u$ and $v$, though you have no specified whether they are real or complex. In the next paragraph, you write $w=x+iy$, so that $w$ is a single number, then write $f(w)$. This contradicts the idea that $f$ was a function of two variables. Your question is fundamentally confused.
Now, assuming $u$ and $v$ are meant to be real numbers, then of course it holds, since you have simply written the definition of the absolute value.
However, if $u$ and $v$ are complex numbers then no, it is false. Just take $u=i$ and $v=1$. Then $f(i, 1)=2i$ so $|f(i,1)|=2$. However, $(2i)^{1/2} \neq 2$ because $2^2=4$ and $(2i)^2 = -4$.
A: Yes. If $f(z) = u(z) + iv(z)$ where $u$ and $v$ are real-valued functions, then
$$
|f(z)| = \sqrt{|u(z)^2 + v(z)^2|},
$$
and so (writing $|f|$ for the function $|\cdot|\circ f$ as usual) you have $|f| = \sqrt{u^2 + v^2}$. Accordingly, we can think of $|f|$ as a function $\mathbb{C}\to\mathbb{C}$ (or $\mathbb{C} \to \mathbb{R}$, or $\mathbb{R}^2 \to \mathbb{R}$, etc.).
One suggestion: it might make your life easier to get into the habit of writing $\sqrt{\alpha}$ to denote the positive square root of $\alpha$ and reserve the notation $\alpha^{1/2}$ for the set of both roots of $\alpha$. Of course, notations and conventions vary, but following a convention like this will make discussing the branches of complex functions easier when you get to that topic.
