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Question Number 130440 by mohammad17 last updated on 25/Jan/21

Commented by mohammad17 last updated on 25/Jan/21

$$helpme$$

Answered by mathmax by abdo last updated on 25/Jan/21

$$\mid \mathrm{z}-1\mid =1\Rightarrow \mathrm{z}-1={\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}\theta \in ]0,2\pi [\Rightarrow \mathrm{z}=1+{\mathrm{e}}^{\mathrm{i}\theta }$$$$\mathrm{z}=1+\cos \theta +\mathrm{isin}\theta ={2\cos }^2\left(\frac{\theta }{2}\right)+2\mathrm{isin}\left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)$$$$=2\cos \left(\frac{\theta }{2}\right){\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}}\mathrm{if}\cos \left(\frac{\theta }{2}\right)>0\Rightarrow \mathrm{argz}=\frac{\theta }{2}\Rightarrow 2\mathrm{argz}=\theta \arg (\mathrm{z}-1)$$$$\frac{2}{3}\arg ({\mathrm{z}}^2-\mathrm{z})=\frac{2}{3}\arg \left(\mathrm{z}(\mathrm{z}-1)\right)=\frac{2}{3}\{\mathrm{argz}+\arg (\mathrm{z}-1)\}$$$$=\frac{2}{3}\{\frac{\theta }{2}+\theta \}=\frac{2}{3}\times \frac{3}{2}\theta =\theta =\arg (\mathrm{z}-1)$$$$\mathrm{we}\mathrm{use}\mathrm{the}\mathrm{same}\mathrm{proof}\mathrm{if}\cos \left(\frac{\theta }{2}\right)<0\mathrm{because}$$$$\mathrm{z}=-2\cos \left(\frac{\theta }{2}\right){\mathrm{e}}^{\mathrm{i}\pi }.{\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}}=-2\cos \left(\frac{\theta }{2}\right){\mathrm{e}}^{\mathrm{i}(\pi +\frac{\theta }{2})}\Rightarrow \mathrm{argz}=\pi +\frac{\theta }{2}$$$$\mathrm{the}\mathrm{equality}\mathrm{is}\mathrm{modulo}2\pi .$$

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