Show that $\left|\dfrac{R(z-a)}{R^{2}-\overline{a}z}\right|$ is analytic for $|z| \leq R$, and maps the circle $|z| = R$ into the unit circle. 
a) Suppose $|a| < R$, show that
  $$\left|\dfrac{R(z-a)}{R^{2}-\overline{a}z}\right|$$ is analytic for $|z| \leq
 R$, and maps the circle $|z| = R$ into the unit circle. 
b) Suppose $|a_k| <R$ for $k=1,2,...,n$. Prove that (unless $|a_k| =
 0$ for all $k$) we have  $$\sqrt[n]{|z-a_1|\cdot|z-a_2|
 \cdot\cdot\cdot |z-a_n|}$$ assumes a maximum value greater than $R$,
  and a minimum value less than $R$ at some points $z$ on $|z| = R$.
   Hint: Consider applying Maximum and Minimum Modulus to
  $\prod_{k=1}^{n}(R^{2}-\overline{a_k}z)$

a) I would assume the question meant that $R>1$. Now to show $f(z) = \left|\dfrac{R(z-a)}{R^{2}-\overline{a}z}\right|$ is analytic in $|z| \leq R$, it suffices to check the denominators of $f$ are outside the range of $|z| = R$. Indeed, the pole $z = \dfrac{R^{2}}{\overline{a}}$ is outside the circle, as one can verify with $$\dfrac{|R^{2}|}{|\overline{a}|} > \dfrac{|R^{2}|}{R} = R$$
Next, we verify that the function $f$ indeed serves as a mapping from $|z| =R$ to $|z|=1$. It suffices to show that for $|a| < R$ and for all $|z| = R$, we have $|f(z)| = 1$. 
To show $|f(z)| =1 1$, it suffices to show $\left|R(z-a)\right|^{2} -|R^{2}-\overline{a}z|^{2} = 0$ on $|z| = R$ and $|a| < R$.
After expanding out, 
$\begin{aligned}
\left|R(z-a)\right|^{2} -|R^{2}-\overline{a}z|^{2}  & = R(z-a)\cdot R(\overline{z}-\overline{a}) - (R^{2}-\overline{a}z)(R^{2}-a\overline{z})\\
  &= R^{2}(z\overline{z}-z\overline{a}-a\overline{z}+a\overline{a})- (R^{4}-R^{2}a\overline{z}-\overline{a}zR^{2}+a\overline{a}z\overline{z})\\
  &= R^{4}-R^{2}\overline{a}z-R^{2}a\overline{z}+R^{2}|a|^{2}-R^{4}+R^{2}a\overline{z}+R^{2}\overline{a}z-|a|^{2}R^{2}\\
  &= R^{4}R^{2}|a|^{2}-R^{4}-|a|^{2}R^{2}\\
  &= R^{4}-R^{4} =0 \\
  &< 0
\end{aligned}$
Hence we have verified part a). 
b) We need to show $$\left|\left(|z-a_1|\cdot |z-a_2| \cdot \cdot \cdot |z-a_n|\right)^{n}\right| < (R+\epsilon_1)$$
On $|z| = R$, we have $$\left|\dfrac{R(z-a_1)\cdot R(z-a_2) \cdot\cdot\cdot R(z-a_n)}{\prod_{k=1}^{n}\left(R^{2}-\overline{a_k}z\right)}\right| = 1 ~~~\text{ by part a) }$$
By simplifying, we have $$|z-a_1|\cdot |z-a_2| \cdot\cdot\cdot |z-a_n| = \left|\dfrac{\prod_{k=1}^{n}\left(R^{2}-\overline{a_k}z\right)}{R^{n}}\right|$$
$\left|\prod_{k=1}^{n}\left(R^{2}-\overline{a_k}z\right)\right| = \left|\prod_{k=1}^{n}R^{2}\left(1-\dfrac{\overline{a_k}z}{R^{2}}\right)\right| \leq R^{2n}\left|\prod_{k=1}^{n}\left(1+\overline{a_k}\right)\right|$
Thus it follows that $$|z-a_1|\cdot |z-a_2| \cdot\cdot\cdot |z-a_n| = \left|\dfrac{\prod_{k=1}^{n}\left(R^{2}-\overline{a_k}z\right)}{R^{n}}\right| \leq R^{n}\left|\prod_{k=1}^{n}\left(1+\overline{a_k}\right)\right|$$

For part $a)$, i am not even sure if i am right, can anyone please
  verify. For part b), i have some clue, but i cannot get to apply the
  max/min modulus principle hence there must be something wrong.
  Furthermore, the question says that it assumes a maximum value greater
  than $R$, however, my proof can only show that it is bounded by some
  bound and not really answering the question.
Please help me, it is from Bak and newman, chapter 7.

 A: Your answer for part A seems to be correct, although there are likely shorter ways to do it. 
For part B, I think the answer you are looking for is here:
Show that $\sqrt[k]{|z-a_1|\cdots |z-a_k|}$ has a max greater than $R$, and a min less than $R$
You don't actually need to convert $z = e^{\theta i} $ form. 
Let $f(z) = \left| z - a _ { 1 } \right| \cdot \left| z - a _ { 2 } \right| \cdots \left| z - a _ { n } \right|$. You want to show that there exists a point in $|z| \leq R$ where $f(z) = R$. For instance, $f(0)$. If you plug $z = 0$ into $\prod _ { k = 1 } ^ { n } \left( R ^ { 2} - \overline { a _ { k } } z \right)$ since they are equivalent expressions, you will find that $|R^n|f(0) = \prod _ { k = 1 } ^ { n }  R ^ { 2 } = R^{2n} \implies f(0) = R^n$. So the function you care about is $g(z) = f(z)^{\frac{1}{n}}$, which means that $g(0) = R$. $z = 0$, however is not the boundary. So by Min/Max Mod Theorem, there must be a minimum or a maximum on the boundary such that it is less than/greater than $R$ and we are done.  
