How find this maximum of this complex numbers of $x,y$ let $x,y$ be complex numbers,such that $|x|=|y|=1$.
Can anyone help me to find the maximum value of the following expression

$$|1+x|+|1+xy|+|1+xy^2|+\cdots+|1+xy^{2013}|-1007|1+y|$$ 

my try: let 
$$x=a+bi, y=c+di$$ then
$$a^2+b^2=1,c^2+d^2=1$$
then I fell very ugly, but I think this problem has nice solutions,becasuse this problem is from china 2013 Mathematical olympiad
 A: Note that $$ |z|=1 \Longrightarrow |1+z|+|1-z|\le 2\sqrt{2} $$
and triangle inequality
$$|1+x|+|1+xy|\le |1+x|+|1+y|+|1-x|\le |1+y|+2\sqrt{2}  $$ 
$$ |1+xy^2|+|1+xy^3|\le |1+xy^2|+|1+y|+|1-xy^2|\le |1+y|+2\sqrt{2} $$.
A: In order for the difference between A and B to be maximal, we need A to be maximal, and B to be minimal. Since B is a module, its minimal value is $0$ , in which case y = $-1$. In order for A to be maximal when y = $-1$, we need the real part of $x$ to be $0$, so that nothing will be subtracted from $1$ when added to $xy^n$ ; in which case, the only two options are y = $\pm\ i$, since |y| = $1$. The only other path of attack would be to start by making A maximal first, in which case x = y = $1$: but this has also the disadvantage of maximizing B as well, yielding a far lesser result than in the former case.
A: For the expression:
$$|1+x|+|1+xy|+|1+xy^2|+\cdots+|1+xy^{2013}|-1007|1+y|$$to be maximum the expression $1007|1+y|$ should be minimum i.e., $|1+y|$ should be minimum.
By modulus properties $|1+y|\ge|{1-|y|}|=|1-1|$ i.e., $|1+y|\ge{0}$. So, the minimum value of $|1+y|=0$.
Let $y=c+di$.
$|1+y|=|(c+1)+di|=\sqrt{(c+1)^2+d^2}=0.$ i.e., $c=-1$ & $d=0$.
 Thus, $$y=-1$$ Substituting it into the expression it becomes $$|1+x|+|1-x|+|1+x|+\cdots+|1-x|=1007|1+x|+1007|1-x|$$ Now, $|x|=|y|=1$ i.e., $|x|^2=|y|^2=1=x\bar{x}=y\bar{y}$. Hence, $$x=\frac{1}{\bar{x}}$$ And, $$y=\frac{1}{\bar{y}}$$ and vice versa.
After this there are two methods.
1st method:
By using the property that if $|z|=1$ then $|1+z|+|1-z|\le2\sqrt{2}$.
Thus, max. value is $$1007[|1+x|+|1-x|]=1007*2\sqrt{2}=2014\sqrt{2}$$
2nd method:
This method is a bit long. Use this if you are not familiar with the above property or cannot prove it. For the expression to be max. it's first derivative is $=0$.
Now use, $|1+x|=\sqrt{(1+x)(1+\bar{x})}$ and $|1-x|=\sqrt{(1-x)(1-\bar{x})}$.
By differentiating the expression twice with respect to x we get it's max. value as $x=\pm{i}$. By substitution it's value in the expression it becomes:
$$1007[|1+i|+|1-i|]=1007*2\sqrt{2}=2014\sqrt{2}$$
