What is radical 3 divided by eight minus 1 divided by eight i expressed in polar form?

First I will state that I am doing this by 'ear' my technique may not be as 'neat' as others may provide.

I think my answer is correct.  :)

 

$$\\\sqrt3\div 8-1\div8i\\\\
\frac{\sqrt3}{8}-\frac{1}{8}i$$

$$\\rcos\theta=\frac{\sqrt3}{8}\qquad rsin\theta=-\frac{1}{8}\qquad r>0\\\\
\frac{rsin\theta}{rcos\theta}=\frac{-1}{8}\div\frac{\sqrt3}{8}\\\\
tan\theta=\frac{-1}{\sqrt3}\\\\
Since\; cos\theta $ is pos and $sin\theta$ is neg $\theta $ must be in the 4th quadrant $\\\\
\theta=\frac{-\pi}{6}\\\\
$Now find r$\\\\
rcos\theta=\frac{\sqrt3}{8}=r\times\frac{\sqrt3}{2}\\\\
\frac{\sqrt3}{2}*\frac{1}{4}=r\times\frac{\sqrt3}{2}\\\\
r=\frac{1}{4}=0.25\\\\
so\\\\$$

 

$$\\\frac{\sqrt3}{8}-\frac{1}{8}i=0.25(cos\frac{11\pi}{6}+isin\frac{11\pi}{6})$$