Let $z_1 = 1 + \sqrt{2}i$ and $z_2 = 1 - \sqrt{2}i$
(a) Determine the polar of $z_1$
(b) Determine the polar of $z_2$
My attempt for (a) looks as follow:
$z_1 = 1 + \sqrt{2}i$
x = 1, y = $\sqrt{2}$
r = $\sqrt{1^2 + \sqrt{2}^2} = \sqrt{3}$
$\theta = tan^{-1}(\dfrac{\sqrt{2}}{1})$
I'm stuck here...I don't know the value for theta from here. It's easier, for me, when I can use cos$\theta$, for example, $cos\theta = \dfrac{1}{2}$, then $\theta = \dfrac{\pi}{2}$. What's the full solution to my problem?
If
$$\tan\theta=t$$ then $$\cos\theta=\pm\frac1{\sqrt{1+t^2}}$$ allows you to convert tangent to cosine. In the case at hand,
$$\tan\theta=\sqrt2\implies \cos\theta=\frac1{\sqrt 3}.$$
Anyway, you'd better train on taking arc tangents in addition to arc cosines.