Question

if z=(cos©,sin©)find z-1/z
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Answer 1

Explanation:

METHOD 1 :

cosθ+isinθ-1/cosθ+i sinθ

=cosθ+i sinθ-(cosθ+i sinθ)^-1 //(cosθ+i sinθ)^-1=cos(-θ)+i sin(-θ)

=cosθ+i sinθ-(cos(-θ)+ i sin(-θ)

=cosθ+i sinθ-(cosθ-isinθ)

ans.=i2sinθ

METHOD 2:

(cosθ+i sinθ)-1/(cosθ+i sinθ)

=((cosθ+i sinθ)^2–1)/cosθ+i sinθ // using LCM

(cosθ+i sinθ)^2–1)*(cosθ-i sinθ)/(cosθ+i sinθ)*(cosθ-isinθ) // using rationalization

=(cosθ+i sinθ)^2–1)*(cosθ-i sinθ) // i^2=-1 and cosθ^2+sinθ^2=1

=(((cosθ)^2+i^2(sinθ)^2+2isinθcosθ or isin2θ)-1)*(cosθ-i sinθ) //(a+b)^2=a^2+b^2+2ab

=(((cosθ)^2-(sinθ)^2+isin2θ)-1)*(cosθ-i sinθ)

=(cos2θ+isin2θ-1)*(cosθ-i sinθ) // cos^2(θ)-sin^2(θ)=cos2θ

by multiplication ,we get

=cosθcos2θ-isinθcos2θ+isin2θcosθ-i^2sinθsin2θ-cosθ+i sinθ

// cos a cos b+sin a sin b=cos(a-b)

cos(2θ-θ)+i(-sinθcos2θ+sin2θcosθ+sinθ)-cosθ

// sin a cos b+cos a sin b=sin(a+b)

=i(sin(2θ-θ)+sinθ)

ans.=i2sinθ

hope will be helpful