Answer
a) $(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) $(4, \dfrac{2\pi}{3}, 3)$
Work Step by Step
In the cylindrical co-ordinates system, we have $x=r \cos \theta$;$ y=r \sin \theta$ and $r^2=x^2+y^2 $
a) Here, $ r=\sqrt 2$
and $x=r \cos \theta \implies -1=\sqrt 2 \cos \theta$
$ \cos \theta=\dfrac{-1}{\sqrt 2} \implies \theta=\dfrac{3\pi}{4}$
So, we have $(r,\theta, z)=(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) Here, $r=\sqrt{(-2)^2+(2\sqrt 3)^2}=\sqrt{4+12}=4$
and $x=r \cos \theta \implies -2=4 \cos \theta$
$\cos \theta =\dfrac{-1}{2} \implies \theta=\dfrac{2\pi}{3}$
so, we have $(r,\theta, z)=(4, \dfrac{2\pi}{3}, 3)$
