Modulus power rule I am not sure whether the following equality holds for complex number:
Let $x$ be complex number and $n$ be positive integer. Then $|x^n|=|x\cdot …\cdot x|=|x|\cdot …\cdot |x|=|x|^n$.
I think the case in real is trivial and I have seen someone tried to exhaust all the 4 cases and had proven that $|x||y|=|xy|$ for $x,y\in\mathbb{R}$.
 A: $[1[.$ Let $a,b,a',b'\in\Bbb R$ and $i^2=-1.$ Let $x=a+bi$ and $x'=a'+b'i.$ Then $xx'=(aa'-bb')+i(ab'+a'b).$ So $$|xx'|^2=(aa'-bb')^2+(ab'+a'b)^2=$$ $$=(a^2a'^2+b^2b'^2-2ab'a'b)+(a^2b'^2+a'^2b^2+2ab'a'b)=$$ $$=a^2a'^2+b^2b'^2+a^2b'^2+a'^2b^2 =$$ $$=(a^2+b^2)(a'^2+b'^2)=$$ $$=|x|^2\cdot |x'|^2.$$ So $|xx'|=|x|\cdot |x'|.$
In particular, if $x'=x$ then we have $|x^2|=|x|^2.$ Hence $|x^3|=|x\cdot x^2|=|x|\cdot |x^2|=|x|\cdot |x|^2=|x|^3.$ Hence $|x^4|=|x\cdot x^3|=|x|\cdot |x^3|=|x|\cdot |x|^3=|x|^4$, et  cetera.
$[2].$ If $u,v\in\Bbb R$ and $u^2+v^2=1$ then there exists $t\in\Bbb R$ such that $(u,v)=(\cos t,\sin t).$ So if $0\ne x=a+bi$ (as above) and $(\frac {a}{|x|},\frac {b}{|x|})=(u,v)$ then there exists $t\in \Bbb R$ such that $(\frac {a}{|x|},\frac {b}{|x|})=(\cos t,\sin t)$ and hence $x=|x|(\cos t+i\sin t).$
Of course if $x=0$ then $|x|=0$ and we can also write $x=0\cdot (\cos t+i\sin t)=|x|(\cos t+i\sin t)$ for some (any) $t.$
Suppose $t,t',t''\in\Bbb R$ and $x=|x|(\cos t+i\sin t)$ and $x'=|x'|(\cos t'+i\sin t')$ and  and $xx'=|xx'|(\cos t''+i\sin t'').$ Then by the Trigonometric Angle-Sum Formulas we have $$|xx'|(\cos t''+i\sin t'')= xx'=$$ $$=|x|\cdot |x'|(\,(\cos t\cos t'-\sin t \sin t')+i(\cos t \sin t'+ \cos t' \sin t)\,)=$$ $$=|x|\cdot |x'|(\cos (t+t')+i\sin (t+t'))$$ and hence $|xx'|=|x|\cdot |x'|.$
