If $x+i y=\frac{3}{2+\cos (\theta)+i \sin (\theta)}$, then $x^2+y^2=$

Question

Aicharya · Accepted Answer

It is given that, $\begin{aligned} x+i y & =\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^2+\sin ^2 \theta} \ \Rightarrow \quad x+i y & =\frac{3(2+\cos \theta)}{5+4 \cos \theta}-i \frac{3 \sin \theta}{5+4 \cos \theta} \ \because \quad x^2+y^2 & =(x+i y)(x-i y)=\frac{9}{(2+\cos \theta)^2+\sin ^2 \theta} \ & =\frac{9}{5+4 \cos \theta} \ \because 4 x-3 & =\frac{12(2+\cos \theta)}{5+4 \cos \theta}-3 \ & =\frac{24+12 \cos \theta-15-12 \cos \theta}{5+4 \cos \theta}=\frac{9}{5+4 \cos \theta} \ \therefore \quad x^2+y^2 & =4 x-3 \end{aligned}$ Hence, option (a) is correct.

If \(x+i y=\frac{3}{2+\cos (\theta)+i \sin (\theta)}\), then \(x^2+y^2=\)

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