Question

If x + iy = cis $\alpha$ .cis $\beta$. Find the value of $x^2+y^2$

Answer 1

Given:- $x+iy=cis\alpha \cdot cis\beta$

To find:- $x^2+y^2$

Solution:-

$x+iy$ $=cis\alpha \cdot cis\beta =(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )=(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+i(\sin \alpha \cos \beta +\cos \alpha \sin \beta )=\cos (\alpha +\beta )+i\sin (\alpha +\beta )$

On comparing both sides, we get

$x=\cos (\alpha +\beta )$

$y=\sin (\alpha +\beta )$

Therefore,

$x^2+y^2$ $={\left(\cos (\alpha +\beta )\right)}^2+{\left(\sin (\alpha +\beta )\right)}^2={\cos }^2(\alpha +\beta )+{\sin }^2(\alpha +\beta )=1$

Hence the value of $x^2+y^2$ is $1$.