Polar form of the sum of complex numbers $\operatorname{cis} 75 + \operatorname{cis} 83 + \ldots+ \operatorname{cis} 147$ 
The number $\operatorname{cis} 75 + \operatorname{cis} 83 + \operatorname{cis} 91 +\dots+ \operatorname{cis} 147$ is expressed in the form $r\operatorname{cis}(\theta)$, where $0\leq \theta< 360$. 
  Find $\theta$ in degrees

I'm having major trouble with this problem.
 A: Hint: you want to evaluate $$e^{i(75\pi/180)}+e^{i(83\pi/180)}+e^{i(91\pi/180)}+\cdots +e^{i(147\pi/180)}.$$
This is a geometric series with common ratio $r=e^{i\frac{8\pi}{180}}.$
Now use the formula for the sum to $n$ terms of a geometric series: $$S_n=\frac{1-r^n}{1-r}.$$

Also, don't forget to convert back to degrees using 
$$\boxed{\theta^\circ =\theta^{\ \rm{rad}}\times \frac{180}{\pi}}.$$
A: Rearrange the 10 terms as:
$$ (\operatorname{cis} 75 + \operatorname{cis} 147) +
(\operatorname{cis} 83 + \operatorname{cis} 139) + \cdots + (\operatorname{cis} 107 + \operatorname{cis} 115)$$
By the parallelogram rule, each pair of terms here is the diagonal of a rhombus which goes in the direction halfway between the two angles -- that is, in this case always parallel to the direction $\frac{75+147}{2} = 111$. All that remains to see is whether the actual direction is $111$ or its opposite.
However, if the angles are in degrees, then we easily see that all of the original terms have positive imaginary part -- and therefore so must their sum, so the answer is $111^\circ$.
