A proof that $|x+yi|=\sqrt{x^2+y^2}$, based on the given the conditions

Question

If we attempt to define $|x+yi|$ by following conditions:

1. $|x|=|xi|=x\operatorname{sgn}(x)$ (implicitly meaning the result will always be $\ge 0$)

2. $|xz|=|x||z|$

3. $|z^x|=|z|^x$

for $x \in \mathbb{R}, z \in \mathbb{C}$. (Note that the definitions of $x$ and $z$ apply to all three conditions listed, not just the last one.)

I want to prove for all real $x$ and $y$ that $|x+yi|=\sqrt{x^2+y^2}$, but proofs that only apply to subsets of all of the possible real values of $x$ and $y$ are welcomed (I show one proof below that proves $|x+xi|=\sqrt{2x^2}$). I've managed to quite easily prove that $|1+i|=\sqrt 2$: $$|1+i|=? \\ |(1+i)^2|=|1+2i-1|=|2i|=|2|=2 \\ |1+i|=\sqrt2$$ However, the moment I start to generalise to $|x+i|$, $|1+xi|$ or $|x+yi|$, I am incapable of getting rid of the imaginary component on the right-hand side. $$|x+i|=?\\|(x+i)^2|=|x^2+2ix-1|$$ From then on, no matter how much I try to manipulate the equation, I can't get rid of the imaginary component on the right hand side, same case with $|x+iy|$ and $|1+xi|$. (I don't know if I'm missing something or if there's some way of manipulating the equation that I don't know about.) I generalised $|1+i|=\sqrt{2}$ to $|x+xi|=\sqrt{2x^2}$ by using the same method I used on $|1+i|$ (by taking advantage of the "mutually annihilative terms" when multiplying out the brackets): $$|x+xi|=? \\ |(x+xi)^2|=|x^2+2x^2i-x^2|=|2x^2i|=|2x^2|=2x^2 \\ |x+xi|=\sqrt{2x^2}$$

[Edit by OP] I recently have found this generalised "polar" form, in case it is of use to proving $|x+yi|=\sqrt{x^2+y^2}$: For $x,y \in \mathbb{R}$, $$|i|=1 \\ |i^y|=1^y=1 \\ |xi^y|=x$$

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Restating the question for clarity:

Assume that the function $f:\Bbb{C}\to\Bbb{R}_{\ge0}$ satisfies the criteria

1. $f(x)=f(ix)=|x|$ for all $x\in\Bbb{R}$,
2. $f(xz)=f(x)f(z)$ for all $x\in\Bbb{R},z\in\Bbb{C}$, and
3. $f(z^x)=f(z)^x$ for all $x\in\Bbb{R},z\in\Bbb{C}$.

Then prove that $f(x+iy)=\sqrt{x^2+y^2}$ for all $x,y\in\Bbb{R}.$

[Edit by OP] For the third condition, note that if the ambiguity with multiple roots become a problem, you may restrict the answer to $\sqrt[x]{z}$ to the root which gives the complex number with the smallest non-negative $\theta$ component when describing complex numbers using the polar coordinate system (as in: $xi^y$).

Answer 1

Yes, as it has been restated now, your question can be answered in the affirmative (ignoring details such as the fact that $z^x$ cannot be defined for just any $x \in \Bbb R$ and $z \in \Bbb C$).

First, assume that $|z|=1$ and $z \ne 1$. Then, there exist $t \in (0, 1)$ such that $z = \textrm e ^{2 \pi \textrm i t}$. Note that

$$z^{\frac 1 t} = \textrm e ^{\frac 1 t \log z} = \textrm e ^{\frac 1 t 2 \pi \textrm i t} = \textrm e ^{2 \pi \textrm i} = 1 ,$$

so from conditions (1) and (3):

$$1 = |1| = f(1) = f(z^\frac 1 t) = f(z) ^{\frac 1 t} ,$$

so $f(z) = 1^t = 1 = |z|$.

If $z = 1$, then $f(z) = f(1) = |1| = |z|$ from condition (1). So far, we have proved that if $|z| = 1$, then $f(z) = 1 = |z|$.

Now, if $z \ne 0$ then $\frac z {|z|}$ has modulus $1$, so we may apply the result from the above paragraph to deduce that $f \left( \frac z {|z|} \right) = 1$, and using condition (2) we obtain

$$f(z) = f \left (|z| \frac z {|z|} \right) = f(|z|) f \left( \frac z {|z|} \right) = |z| \cdot 1 = |z| .$$

Finally, if $z=0$ then condition (1) gives $f(0) = 0 = |0|$ trivially.

We have thus proved that $f(z) = |z| \ \forall z \in \Bbb C$.

(Note that the part $f(\textrm i x) = |x|$ from condition (1) was not needed, it can be deduced from (3) and the other part of (1).)