If z = cos ⁡ θ + i sin ⁡ θ prove z 2 − 1 z 2 + 1 = i tan ⁡ θHere is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated. ( cos ⁡ θ + i sin ⁡ θ ) 2 − 1 ( cos ⁡ θ + i sin ⁡ θ ) 2 + 1 ( cos 2 ⁡ θ + 2 i sin ⁡ θ cos ⁡ θ − sin 2 ⁡ θ ) − 1 ( cos 2 ⁡ θ + 2 i sin ⁡ θ cos ⁡ θ − sin 2 ⁡ θ ) + 1 ( cos 2 ⁡ θ − sin 2 ⁡ θ ) + ( 2 i sin ⁡ θ cos ⁡ θ ) − 1 ( cos 2 ⁡ θ − sin 2 ⁡ θ ) + ( 2 i sin ⁡ θ cos ⁡ θ ) + 1 cos ⁡ 2 θ + i sin ⁡ 2 θ − 1 cos ⁡ 2 θ + i sin ⁡ 2 θ + 1 I understand how I can do it with using z = e i θ , however I want to solve it using double angle identities.

Question

If   z  =  cos  ⁡  θ  +  i  sin  ⁡  θ  prove                     z        2            −      1                      z        2            +      1        =  i  tan  ⁡  θHere is my workings, I&#039;m not sure if I&#039;ve made a mistake or I&#039;m just not spotting what to do next. Any help would be appreciated.            (      cos      ⁡      θ      +      i      sin      ⁡      θ              )        2            −      1              (      cos      ⁡      θ      +      i      sin      ⁡      θ              )        2            +      1                  (              cos        2            ⁡      θ      +      2      i      sin      ⁡      θ      cos      ⁡      θ      −              sin        2            ⁡      θ      )      −      1              (              cos        2            ⁡      θ      +      2      i      sin      ⁡      θ      cos      ⁡      θ      −              sin        2            ⁡      θ      )      +      1                  (              cos        2            ⁡      θ      −              sin        2            ⁡      θ      )      +      (      2      i      sin      ⁡      θ      cos      ⁡      θ      )      −      1              (              cos        2            ⁡      θ      −              sin        2            ⁡      θ      )      +      (      2      i      sin      ⁡      θ      cos      ⁡      θ      )      +      1                  cos      ⁡      2      θ      +      i      sin      ⁡      2      θ      −      1              cos      ⁡      2      θ      +      i      sin      ⁡      2      θ      +      1      I understand how I can do it with using   z  =      e          i      θ      , however I want to solve it using double angle identities.

jcholewa39v8f · Accepted Answer

Your approach will work with double angle formulae, but this is quicker: since   z  =  exp  ⁡  i  θ,            z      −      1              /            z              z      +      1              /            z        =            2      i      sin      ⁡      θ              2      cos      ⁡      θ

Kaiden Wilkins · Accepted Answer

z  =      e          i      θ       ,                     z        2            −      1                      z        2            +      1        =                    e                  2          i          θ                    −      1                      e                  2          i          θ                    +      1        =                    e                  i          θ                    −              e                  −          i          θ                                    e                  i          θ                    +              e                  i          θ                      =            2      i            s      i      n      (      θ      )              2            c      o      s      (      θ      )        =  i  t  a  n  (  θ  )

If z = cos ⁡ θ + i sin ⁡ &#x

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