Question

If z = (cos Φ,sinΦ), find (z-1/z)
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Answer 1

Answer:

if z is a complex number we can write z=cosθ+i sinθ

We can solve this question by using two methods.There are no difference between both the methods but basic formula’s replacements.

first method is based on complex numbers property.

second method is based on trigonometric formulas.

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.

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