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Question Number 96021 by SmNayon11 last updated on 29/May/20

if p and q are two complex number  and p×q=m  ,m is a real number .    is there always exists a  p^(1/3)  and q^(1/3)   (we know p^(1/3)  and q^(1/3) each has actually   3 values)  such that p^(1/3) ×q^(1/3) =m^(1/3) .where m^(1/3)   is real .??  how to prove it?

$$\mathrm{if}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{p}×\mathrm{q}=\mathrm{m}\:\:,\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:. \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{there}\:\mathrm{always}\:\mathrm{exists}\:\mathrm{a}\:\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{we}\:\mathrm{know}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{each}\:\mathrm{has}\:\mathrm{actually}\:\right. \\ $$$$\left.\mathrm{3}\:\mathrm{values}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} ×\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{m}^{\frac{\mathrm{1}}{\mathrm{3}}} .\mathrm{where}\:\mathrm{m}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{is}\:\mathrm{real}\:.??\:\:\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}? \\ $$$$ \\ $$

Commented by mr W last updated on 29/May/20

p,q∈C  m∈R  pq=m  ⇒(pq)^(1/3) =m^(1/3)   ⇒p^(1/3) q^(1/3) =m^(1/3) ∉R  example:  p=i, q=−i ⇒pq=1=m  p^(1/3) ,q^(1/3) ∈C  but m^(1/3) =third roots of unit, ∉R

$${p},{q}\in\mathbb{C} \\ $$$${m}\in\mathbb{R} \\ $$$${pq}={m} \\ $$$$\Rightarrow\left({pq}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={m}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow{p}^{\frac{\mathrm{1}}{\mathrm{3}}} {q}^{\frac{\mathrm{1}}{\mathrm{3}}} ={m}^{\frac{\mathrm{1}}{\mathrm{3}}} \notin\mathbb{R} \\ $$$${example}: \\ $$$${p}={i},\:{q}=−{i}\:\Rightarrow{pq}=\mathrm{1}={m} \\ $$$${p}^{\frac{\mathrm{1}}{\mathrm{3}}} ,{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \in\mathbb{C} \\ $$$${but}\:{m}^{\frac{\mathrm{1}}{\mathrm{3}}} ={third}\:{roots}\:{of}\:{unit},\:\notin{R} \\ $$

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