Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 67431 by aliesam last updated on 27/Aug/19

Commented by mathmax by abdo last updated on 27/Aug/19

if a and b from C the question is done by sir mjs  if a and b from R   we have  (∣(a/b)∣)^2 −(((∣a∣)/(∣b∣)))^2    =∣(a^2 /b^2 )∣−((∣a^2 ∣)/(∣b∣^2 )) =(a^2 /b^2 )−(a^2 /b^2 ) =0 ⇒(∣(a/b)∣)^2 =(((∣a∣)/(∣b∣)))^(2 )  ⇒  (√((∣(a/b)∣)^2 ))=(√((((∣a∣)/(∣b∣)))^2 )) ⇒∣∣(a/b)∣∣ =∣((∣a∣)/(∣b∣))∣  but the quotient are ≥0 ⇒  ∣(a/b)∣=((∣a∣)/(∣b∣))

$${if}\:{a}\:{and}\:{b}\:{from}\:{C}\:{the}\:{question}\:{is}\:{done}\:{by}\:{sir}\:{mjs} \\ $$$${if}\:{a}\:{and}\:{b}\:{from}\:{R}\:\:\:{we}\:{have}\:\:\left(\mid\frac{{a}}{{b}}\mid\right)^{\mathrm{2}} −\left(\frac{\mid{a}\mid}{\mid{b}\mid}\right)^{\mathrm{2}} \: \\ $$$$=\mid\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\mid−\frac{\mid{a}^{\mathrm{2}} \mid}{\mid{b}\mid^{\mathrm{2}} }\:=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\mathrm{0}\:\Rightarrow\left(\mid\frac{{a}}{{b}}\mid\right)^{\mathrm{2}} =\left(\frac{\mid{a}\mid}{\mid{b}\mid}\right)^{\mathrm{2}\:} \:\Rightarrow \\ $$$$\sqrt{\left(\mid\frac{{a}}{{b}}\mid\right)^{\mathrm{2}} }=\sqrt{\left(\frac{\mid{a}\mid}{\mid{b}\mid}\right)^{\mathrm{2}} }\:\Rightarrow\mid\mid\frac{{a}}{{b}}\mid\mid\:=\mid\frac{\mid{a}\mid}{\mid{b}\mid}\mid\:\:{but}\:{the}\:{quotient}\:{are}\:\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\mid\frac{{a}}{{b}}\mid=\frac{\mid{a}\mid}{\mid{b}\mid} \\ $$

Answered by MJS last updated on 27/Aug/19

a, b ∈C  r, s ∈R^+   α, β ∈R  a=re^(iα)  ⇒ ∣a∣=r  b=se^(iβ)  ⇒ ∣b∣=s  (a/b)=(r/s)e^(i(α−β))  ⇒ ∣(a/b)∣=(r/s)  ⇒ ∣(a/b)∣=(r/s)=((∣a∣)/(∣b∣))

$${a},\:{b}\:\in\mathbb{C} \\ $$$${r},\:{s}\:\in\mathbb{R}^{+} \\ $$$$\alpha,\:\beta\:\in\mathbb{R} \\ $$$${a}={r}\mathrm{e}^{\mathrm{i}\alpha} \:\Rightarrow\:\mid{a}\mid={r} \\ $$$${b}={s}\mathrm{e}^{\mathrm{i}\beta} \:\Rightarrow\:\mid{b}\mid={s} \\ $$$$\frac{{a}}{{b}}=\frac{{r}}{{s}}\mathrm{e}^{\mathrm{i}\left(\alpha−\beta\right)} \:\Rightarrow\:\mid\frac{{a}}{{b}}\mid=\frac{{r}}{{s}} \\ $$$$\Rightarrow\:\mid\frac{{a}}{{b}}\mid=\frac{{r}}{{s}}=\frac{\mid{a}\mid}{\mid{b}\mid} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com