Question Number 130599 by mohammad17 last updated on 27/Jan/21
$${by}\:{using}\:{diffintion}\:{lim}_{{z}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{\mid{z}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$$${help}\:{me}\:{sir} \\ $$
Answered by mathmax by abdo last updated on 27/Jan/21
$$\mathrm{z}=\mathrm{e}^{\mathrm{i}\theta} \:\:\:\left(\mathrm{z}\rightarrow\mathrm{1}\:\Rightarrow\theta\:\rightarrow\mathrm{0}\:\mathrm{and}\:\mathrm{z}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{e}^{\mathrm{2i}\theta} −\mathrm{1}\:=\mathrm{cos}\left(\mathrm{2}\theta\right)+\mathrm{isin}\left(\mathrm{2}\theta\right)−\mathrm{1}\right. \\ $$$$=−\mathrm{sin}^{\mathrm{2}} \left(\theta\right)+\mathrm{2isin}\theta\:\mathrm{cos}\theta\:\Rightarrow\mid\mathrm{z}^{\mathrm{2}} −\mathrm{1}\mid=\mid\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{2icos}\theta\mathrm{sin}\theta\mid \\ $$$$=\mid\mathrm{sin}\theta\left(\mathrm{sin}\theta−\mathrm{icos}\theta\right)\mid=\mid\mathrm{sin}\theta\mid\:\:\Rightarrow\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{\mid\mathrm{z}^{\mathrm{2}} −\mathrm{1}\mid}=\mathrm{lim}_{\theta\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}}{\mid\mathrm{sin}\theta\mid}=+\infty \\ $$