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Question Number 143097 by ZiYangLee last updated on 10/Jun/21

If z=cos θ+isin θ, prove that ((1−z^2 )/(1+z^2 ))=−itan θ

$$\mathrm{If}\:{z}=\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta,\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{2}} }=−{i}\mathrm{tan}\:\theta \\ $$

Answered by MJS_new last updated on 10/Jun/21

((1−(c+is)^2 )/(1+(c+is)^2 ))=((1−c^2 +s^2 −2ics)/(1−s^2 +c^2 +2ics))=((s^2 −ics)/(c^2 +ics))= =((s(s−ic))/(c(c+is)))=((−isz)/(cz))=−i tan θ

$$\frac{\mathrm{1}−\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{c}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} −\mathrm{2ics}}{\mathrm{1}−\mathrm{s}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{2ics}}=\frac{\mathrm{s}^{\mathrm{2}} −\mathrm{ics}}{\mathrm{c}^{\mathrm{2}} +\mathrm{ics}}= \\ $$$$=\frac{\mathrm{s}\left(\mathrm{s}−\mathrm{ic}\right)}{\mathrm{c}\left(\mathrm{c}+\mathrm{is}\right)}=\frac{−\mathrm{is}{z}}{\mathrm{c}{z}}=−\mathrm{i}\:\mathrm{tan}\:\theta \\ $$

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