C'mon. If you are at this level you are betting than this. Apply yourself! To find a max or min you take the derivative and set it equal to 0. As for the function:
$\displaystyle |1 + Re^{i \theta}| = |(1 + R~cos(\theta)) + i~R~sin(\theta)| = \sqrt{(1 + R~cos(\theta))^2 + (R~sin(\theta))^2}$
$\displaystyle = \sqrt{1 + 2R~cos(\theta) + R^2~cos^2(\theta) + R^2~sin^2(\theta)} = \sqrt{1 + 2R~cos(\theta) + R^2}$
So, for example, you are trying to find the minimum of $\displaystyle |1 + Re^{i \theta}|$:
$\displaystyle \frac{d}{d \theta}|1 + Re^{i \theta}| = - \frac{R~sin(\theta)}{\sqrt{1 + 2R~cos(\theta) + R^2}}$
Setting this equal to 0 gives $\displaystyle \theta = n \pi$ or R = 0. (If all you care about is the variation of (theta) then ignore the R = 0 possibility, though it really can potentially affect the final answer.)
Use the second derivative test to decide if n is even or odd. The proof using the maximum is not much more complicated.
-Dan
$\displaystyle 1+ Re^{i\theta}$, as $\displaystyle \theta$ goes from 0 to $\displaystyle 2\pi$, is a circle with center at 1= (1, 0) in the complex plane and radius R. The max and min will be at the points closest to and farthest from 0= (0, 0). And those will lie on the straight line through (0, 0) and (1, 0).