Show that for a complex number z

(z+1z)=2cosθ
We know thatz=(cosθ+isinθ)and that1z=(cosθ+isinθ)1Also, by deMoivres Theorem(cosθ+isinθ)1=(cos(θ)+isin(θ))Therefore(z+1z)=(cosθ+isinθ)+(cos(θ)+isin(θ))And becausecos(θ)=cos(θ)andsin(θ)=sin(θ))(z+1z)=(cosθ+isinθ)+(cos(θ)isin(θ))=2cosθ



Show that for a complex number z

(z1z)=2isinθ
We know thatz=(cosθ+isinθ)and that1z=(cosθ+isinθ)1Also, by deMoivres Theorem(cosθ+isinθ)1=(cos(θ)+isin(θ))Therefore(z1z)=(cosθ+isinθ)(cos(θ)+isin(θ))And becausecos(θ)=cos(θ)andsin(θ)=sin(θ))(z1z)=(cosθ+isinθ)(cos(θ)isin(θ))=2isinθ