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Question Number 58592 by rahul 19 last updated on 25/Apr/19

If ∣z−1∣=1, then prove that arg(z) =   (1/2)arg(z−1).

$${If}\:\mid{z}−\mathrm{1}\mid=\mathrm{1},\:{then}\:{prove}\:{that}\:{arg}\left({z}\right)\:=\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{arg}\left({z}−\mathrm{1}\right). \\ $$

Answered by tanmay last updated on 26/Apr/19

z=re^(iθ) =rcosθ+irsinθ  z−1=(rcosθ−1)+irsinθ  ∣z−1∣=(√((rcosθ−1)^2 +(rsinθ)^2 ))  r^2 cos^2 θ−2rcosθ+1+r^2 sin^2 θ=1  r^2 −2rcosθ=0  r(r−2cosθ)=0  r≠0  r=2cosθ  z=2cosθ×cosθ+i2cosθsinθ  A+iB=(2cos^2 θ)+i(2sinθcosθ)  arg(z)=θ  now  z−1  2cos^2 θ+isin2θ−1  =cos2θ+isin2θ  (1/2)arg(z−1)  (1/2)×2θ=θ

$${z}={re}^{{i}\theta} ={rcos}\theta+{irsin}\theta \\ $$$${z}−\mathrm{1}=\left({rcos}\theta−\mathrm{1}\right)+{irsin}\theta \\ $$$$\mid{z}−\mathrm{1}\mid=\sqrt{\left({rcos}\theta−\mathrm{1}\right)^{\mathrm{2}} +\left({rsin}\theta\right)^{\mathrm{2}} } \\ $$$${r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−\mathrm{2}{rcos}\theta+\mathrm{1}+{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{rcos}\theta=\mathrm{0} \\ $$$${r}\left({r}−\mathrm{2}{cos}\theta\right)=\mathrm{0} \\ $$$${r}\neq\mathrm{0}\:\:{r}=\mathrm{2}{cos}\theta \\ $$$${z}=\mathrm{2}{cos}\theta×{cos}\theta+{i}\mathrm{2}{cos}\theta{sin}\theta \\ $$$${A}+{iB}=\left(\mathrm{2}{cos}^{\mathrm{2}} \theta\right)+{i}\left(\mathrm{2}{sin}\theta{cos}\theta\right) \\ $$$${arg}\left({z}\right)=\theta \\ $$$${now} \\ $$$${z}−\mathrm{1} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \theta+{isin}\mathrm{2}\theta−\mathrm{1} \\ $$$$={cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{arg}\left({z}−\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\theta=\theta \\ $$

Commented by JDamian last updated on 25/Apr/19

There is a mistake in the fourth line:  rsin^2 θ    should be    r^2 sin^2 θ

$${There}\:{is}\:{a}\:{mistake}\:{in}\:{the}\:{fourth}\:{line}: \\ $$$${rsin}^{\mathrm{2}} \theta\:\:\:\:{should}\:{be}\:\:\:\:\boldsymbol{{r}}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$

Commented by tanmay last updated on 26/Apr/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by rahul 19 last updated on 26/Apr/19

thank U sir.

$${thank}\:{U}\:{sir}. \\ $$

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