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Question Number 130599 by mohammad17 last updated on 27/Jan/21

b y u s i n g d i f f i n t i o n {l i m}_{z \to 1} \frac{1}{∣ z^{2} - 1 ∣}

h e l p m e s i r

Answered by mathmax by abdo last updated on 27/Jan/21

z = e^{i θ} (z \to 1 \Rightarrow θ \to 0 and z^{2} - 1 = e^{2 i θ} - 1 = \cos (2 θ) + isin (2 θ) - 1

= - \sin^{2} (θ) + 2 isin θ \cos θ \Rightarrow∣ z^{2} - 1 ∣=∣ \sin^{2} θ - 2 icos θ \sin θ ∣

=∣ \sin θ (\sin θ - icos θ) ∣=∣ \sin θ ∣ \Rightarrow \lim_{z \to 1} \frac{1}{∣ z^{2} - 1 ∣} = \lim_{θ \to 0} \frac{1}{∣ \sin θ ∣} = + \infty