Prove-that-z-2-1-z-2-1-i-tan-where-z-cos-i-sin-

Question Number 54543 by Tawa1 last updated on 05/Feb/19

Prove that : \frac{z^{2} - 1}{z^{2} + 1} = i \tan (θ)

where z = \cos (θ) + i \sin (θ)

Commented by maxmathsup by imad last updated on 06/Feb/19

w e h a v e z = e^{i θ} \Rightarrow \frac{z^{2} - 1}{z^{2} + 1} = \frac{e^{2 i θ} - 1}{e^{2 i θ} + 1} = \frac{e^{i θ} - e^{- i θ}}{e^{i θ} + e^{- i θ}} = i \frac{\frac{e^{i θ} - e^{- i θ}}{2 i}}{\frac{e^{i θ} + e^{- i θ}}{2}} = i \frac{s i n θ}{c o s θ} = i t a n θ

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

\frac{z^{2} - 1}{z^{2} + 1}

= \frac{z - \frac{1}{z}}{z + \frac{1}{z}}

z = c o s θ + i s i n θ = e^{i θ}

\frac{1}{z} = \frac{1}{e^{i θ}} = e^{- i θ} = c o s (- θ) + i s i n (- θ) = c o s θ - i s i n θ

s o \frac{z - \frac{1}{z}}{z + \frac{1}{z}} = \frac{2 i s i n θ}{2 c o s θ} = i t a n θ

Commented by Tawa1 last updated on 05/Feb/19

God bless you sir

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