Complex Analysis proof problem I have an feeling this bad proof because I am assuming my conclusion and I go into repeating. This isn't allowed in mathematics, so I need help to refix my steps.
So the proof I am proving is |ab| = |a||b|
So I started with:
Let a,b $\in \mathbb{C}$
I claimed $|ab|^{2} = |a|^{2} |b|^{2}$ is true
so I went to from LHS = RHS (to say LHS is equal to RHS) by using modulus of complex which is $|z|^{2}$ = $zz^{*}$
$|ab|^{2} = aa^{*} (bb^{*})$ = $(ab)(ab)^{*}$ = $|ab|^{2}$
then i took the square root of $|ab|^{2} = |a|^{2} |b|^{2}$ to give me |ab| = |a||b| because modulus is never negative, and we only take the positive.
this is where I believe this proof is 100% written wrong because I am assuming my claim is true (unproven to be exact) and it repeats again. How do I fix this?
 A: Write those numbers in polar form, that is
$$
a=re^{i\theta}\\
b=se^{i\phi}\;.
$$
Since
$$
ab=rse^{i(\theta+\phi)}
$$
and $|a|=r$, $|b|=s$ and $|ab|=rs$, you reach the conclusion easily
A: $|ab|^{2}=(ab)(ab)^{*}=aba^{*}b^{*}=(aa^{*}) (bb^{*})=|a|^{2}|b|^{2}$. Now take square root.
A: You are not assuming what you are trying to prove.  You are assuming two other things.
You are assuming:

*

*That $(mn)^2 = m^2n^2$ for all $m,n\in \mathbb C$.  Which I believe is a valid assumption.

ANd that


*If $x,y$ are non negative reals then $x^2 = y^2 \iff x=y$ which is valid presuming it was proven earlier that $f:[0,\infty)\to [0,\infty)$ via $f(x) = x^2$ is an injective function (presumably it was so proven by proving $f(x)$ is strictly increasing).

Then $|ab|^2 = (|a||b|)^2 = |a|^2|b|^2$ does imply that $|ab| = |a||b|$.
I think you are confusing that you seem to think defining $\sqrt {m^2} = |m|$ somehow makes $\sqrt{(ab)^2} = |ab|$ seem to assume $|ab| =|a||b|$. But it doesn't.
You are assuming $|ab|$ and $|a|$ and $|b|$ are non-negative so $\sqrt{|ab|)^2} = |ab|$ and not necessarily $\big ||ab|\big|$ and $\sqrt{|a|^2|b|^2} = |a||b|$ and not $\big||a||b|\big |$.
If you had concluded that $\sqrt{|ab|^2} =\big ||ab|\big|$ and that $\sqrt{|a|^2|b|^2} =\big||a||b|\big |$ then we would have $\big ||ab|\big|=\big||a||b|\big |$ and you'd have gotten nowher as you'd have no way to prove $\big||a||b|\big |=\big||a|\big |\big||b|\big|$.
But that's moot as we are dealing the strictly non-negative numbers and proving $\sqrt{|ab|^2} = |ab|=\sqrt{|a|^2|b|^2} = |a||b|$ is .... proving $\sqrt{|ab|^2} = |ab|=\sqrt{|a|^2|b|^2} = |a||b|$
