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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 $x,y\in\mathbb{R}$.

sym sym
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  • Use $|xy|=|x||y|$ and induction. Dunno what you mean by four cases. – anon May 08 '22 at 04:13
  • I mean $|xy|=|x||y|$ is true for real $x,y$. But dk whether it is true for complex $x,y$ too. I am not sure how to use induction. They are not discrete numbers. – sym sym May 08 '22 at 04:15
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    @symsym Prove $\vert xy \vert = \vert x \vert \vert y \vert$ for complex $x$ and $y$ then use induction on $n$. It's not that difficult so give it a go. – CyclotomicField May 08 '22 at 04:21
  • Oh I got what you mean now. Thanks – sym sym May 08 '22 at 05:01
  • I stuck at $|x^k||x|$.

    I let $x=re^{i\theta}$ and then $x^k=r^ke^{ik\theta}$.

    Then $|x^k||x|=|r^k||e^{ik\theta}||r||e^{i\theta}|=|r^{k+1}||e^{ik\theta}||e^{i\theta}|$

    Basically I am not sure whether $|x||y|=|xy|$ here. So I can’t assume it is equals to $|x^{k+1}|$.

     – sym sym May 08 '22 at 05:07

1 Answers1

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$[1[.$ 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 $|xx'|=|x|\cdot |x'|.$

DanielWainfleet
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