Just a complex Complex product

  • Thread starter Dinheiro
  • Start date
  • #1
56
0

Homework Statement


Find x
[itex] 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}}) [/itex]


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
 
Last edited:

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,365
1,033

Homework Statement


Find x
[itex] 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}}) [/itex]


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?
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


Find x
[itex] 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}}) [/itex]


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##?
 
  • #4
56
0
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)...
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,365
1,033
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.
 
Last edited:
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
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
[tex] \frac{w-1}{w}\times \left(1+\frac{1}{w} \right) = 1-\frac{1}{w^2} \equiv \frac{w^2-1}{w^2}[/tex]
is helpful here.
 
  • #7
56
0
Thanks, Ray V and SammyS!!
 

Related Threads on Just a complex Complex product

  • Last Post
Replies
5
Views
460
  • Last Post
Replies
1
Views
785
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
807
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
4
Views
866
  • Last Post
Replies
14
Views
992
  • Last Post
Replies
2
Views
822
M
Top