Just a complex Complex product

[Dinheiro]

Homework Statement

Find x
$x = (1 + \frac{1+i}{2})(1 + (\frac{1+i}{2})^{2})(1 + (\frac{1+i}{2})^{2^{2}})...(1 + (\frac{1+i}{2})^{2^{n}})$

Homework Equations

Complex algebra equations

The Attempt at a Solution

Me again and another olympic question. I've tried some trigonometric substitutions, even expand; yet, nothing

 

[SammyS]

Homework Statement

Find x
$x = (1 + \frac{1+i}{2})(1 + (\frac{1+i}{2})^{2})(1 + (\frac{1+i}{2})^{2^{2}})...(1 + (\frac{1+i}{2})^{2^{n}})$

Homework Equations

Complex algebra equations

The Attempt at a Solution

Me again and another olympic question. I've tried some trigonometric substitutions, even expand; yet, nothing

How about showing some of those results?

 

[Ray Vickson]

Homework Statement

Find x
$x = (1 + \frac{1+i}{2})(1 + (\frac{1+i}{2})^{2})(1 + (\frac{1+i}{2})^{2^{2}})...(1 + (\frac{1+i}{2})^{2^{n}})$

Homework Equations

Complex algebra equations

The Attempt at a Solution

Me again and another olympic question. I've tried some trigonometric substitutions, even expand; yet, nothing

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

 

[Dinheiro]

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

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)...

 

[SammyS]

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)...

Did you do what Ray V. suggested?

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

 

[Ray Vickson]

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)...

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

 

[Dinheiro]

Thanks, Ray V and SammyS!