I've solved it by taking modulus of each side. Is there any other possible way to solve this question?

Question

If $\left(a,b,c,d\right) \in \mathbb{R}^4$ and if $$a+bi=\frac{c}{d+\cos(x)+i\sin(x)}$$

Show that $$(d^2-1)(a^2+b^2) +c^2 = 2acd$$

Answer 1

cross multiplication leads to the equation$$ad+a\cos(x)-b\sin(x)+i(bd+b\cos(x)+a\sin(x))=c$$ your equation is true writing $$a\cos(x)-b\sin(x)=-ad$$ and $$a\sin(x)+b\cos(x)=-bd$$ squaring both equations and adding we get $$a^2(\sin^2(x)+\cos^2(x))+b^2(\sin^2(x)+\cos^2(x))=a^2d^2+b^2d^2$$ can you finish? from this equation we get $$a^2+b^2=d^2(a^2+b^2)$$ thus $$d^2=1$$ if $$a^2+b^2\neq 0$$