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Question-67431




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

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