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

$$If\mid z-1\mid =1,thenprovethatarg\left(z\right)=$$$$\frac{1}{2}arg(z-1).$$

Answered by tanmay last updated on 26/Apr/19

$$z={re}^{i\theta }=rcos\theta +irsin\theta$$$$z-1=(rcos\theta -1)+irsin\theta$$$$\mid z-1\mid =\sqrt{{(rcos\theta -1)}^2+{\left(rsin\theta \right)}^2}$$$$r^2{cos}^2\theta -2rcos\theta +1+r^2{sin}^2\theta =1$$$$r^2-2rcos\theta =0$$$$r(r-2cos\theta )=0$$$$r\ne 0r=2cos\theta$$$$z=2cos\theta \times cos\theta +i2cos\theta sin\theta$$$$A+iB=\left(2{cos}^2\theta \right)+i\left(2sin\theta cos\theta \right)$$$$arg\left(z\right)=\theta$$$$now$$$$z-1$$$$2{cos}^2\theta +isin2\theta -1$$$$=cos2\theta +isin2\theta$$$$\frac{1}{2}arg(z-1)$$$$\frac{1}{2}\times 2\theta =\theta$$

Commented by JDamian last updated on 25/Apr/19

$$Thereisamistakeinthefourthline:$$$${rsin}^2\theta shouldbe{\mathit{r}}^2{sin}^2\theta$$

Commented by tanmay last updated on 26/Apr/19

$$thankyousir$$

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

$$thankUsir.$$

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