Just a complex Complex product

1. Aug 3, 2014

Dinheiro

1. The problem statement, all variables and given/known data
Find x
$x = (1 + \frac{1+i}{2})(1 + (\frac{1+i}{2})^{2})(1 + (\frac{1+i}{2})^{2^{2}})...(1 + (\frac{1+i}{2})^{2^{n}})$

2. Relevant equations
Complex algebra equations

3. The attempt at a solution
Me again and another olympic question. I've tried some trigonometric substitutions, even expand; yet, nothing

Last edited: Aug 3, 2014
2. Aug 3, 2014

SammyS

Staff Emeritus
How about showing some of those results?

3. Aug 3, 2014

Ray Vickson

For $z = (1+i)/2$, look at $z^2$. What does this tell you about $z^4, z^8, \ldots$?

4. Aug 3, 2014

Dinheiro

I tried it, though, but I couldn't really solve the recurrency because of the +1 on each factor:
Resolving powers
(1 + (1+i)/2)(1 + i/2)(1 - 1/4)(1 + 1/8)(1 + 1/64)(1 + 1/4096)...
Let's say
a = (1 + (1+i)/2)(1 + i/2)(1 - 1/4)
and
b = (1 + 1/8)(1 + 1/64)(1 + 1/4096)...
We can calculate b's denominator which is
(2^3)[2^(3.2)][2^(3.2.2)][2^(3.2.2.2)]...[2^(3.2.2.2...(n-3))] = 2^[3.2^(n-2) -1]
but b's numerator, I don't know
(9)(65)(4097)...

5. Aug 3, 2014

SammyS

Staff Emeritus
Did you do what Ray V. suggested?

He did not suggest expanding out each factor, just parts of a few specific factors.

Last edited: Aug 4, 2014
6. Aug 4, 2014

Ray Vickson

The simple relation
$$\frac{w-1}{w}\times \left(1+\frac{1}{w} \right) = 1-\frac{1}{w^2} \equiv \frac{w^2-1}{w^2}$$