How to show that |Z|^n=|Z^n|?

Question

I can show that |Z|^2=|Z^2| or |Z|^3=|Z^3| where Z is a complex (a+ib) number. But is there any way of showing that for all n where n is a natural number?

Answer 1

You can really prove a wider result: $|z_1z_2\cdots z_n|=|z_1||z_2|\cdots|z_n|$. (Your statement then follows by setting $z_1=z_2=\ldots=z_n=z$.)

Proof: Hope you know the fact that $|zw|=|z||w|$ for any two complex numbers $z, w$. We can proceed by using induction on $n$:

• $n=1$: trivial because both sides are just $|z_1|$.
• Induction step ($n\to n+1$): $|z_1z_2\cdots z_nz_{n+1}|=|z_1z_2\cdots z_n||z_{n+1}|\text{ (per above fact: take }z=z_1z_2\cdots z_n\text{ and }w=z_{n+1}\text{) }=|z_1||z_2|\cdots|z_n||z_{n+1}|\text{ (inductive hypothesis)}$.

Answer 2

If $w=a+bi$ and $z=c+di$ with $a,\,b,\,c,\,d\in\Bbb R$, $$|wz|^2=(ac-bd)^2+(ad+bc)^2(a^2+b^2)(c^2+d^2)=|w|^2|z|^2.$$ Hence $|wz|=|w||z|$. Now we can induct on $n$.

Answer 3

Since you know that $|z|^2=|z^2|,$ it follows that you must have known that $|w||z|=|wz|.$ Now multiply both sides of the first equation by $|z|,$ to get $$|z|^3=|z^2||z|.$$ Now apply the second equation on RHS to obtain $|z|^3=|z^2||z|=|z^3|.$

Then repeat as many times as you wish.