Question

Compute the product of \[\left[ {1 + \left( {\dfrac{{1 + i}}{2}} \right)} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^2}} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^2}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^n}}}} \right]\], where $n \geqslant 2$

Answer 1

Hint: Here observe the question carefully, it is a product of addition of complex numbers where it continues till n. Therefore here consider a particular term and then observe whether it follows any pattern, so that we can identify the pattern and apply it to the process of solving and finally could end up getting the correct result.

Complete step by step answer:
Consider the first term here \[\left[ {1 + \left( {\dfrac{{1 + i}}{2}} \right)} \right]\], in this term now just consider \[\left( {\dfrac{{1 + i}}{2}} \right)\]:
$ \Rightarrow \dfrac{{1 + i}}{2} = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right)$
Now we know that from Euler’s formula:
$ \Rightarrow {e^{i\theta }} = \sin \theta + i\cos \theta $
$ \Rightarrow \cos {45^ \circ } + i\sin {45^ \circ } = {e^{i\dfrac{\pi }{4}}}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }} = {e^{i\dfrac{\pi }{4}}}$
Now substituting this in the \[\left( {\dfrac{{1 + i}}{2}} \right)\] expression:
$ \Rightarrow \dfrac{{1 + i}}{2} = \dfrac{1}{{\sqrt 2 }}{e^{i\dfrac{\pi }{4}}}$
In the next term there is \[{\left( {\dfrac{{1 + i}}{2}} \right)^2}\], now consider this:
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^2} = {\left( {\dfrac{1}{{\sqrt 2 }}{e^{i\dfrac{\pi }{4}}}} \right)^2}\], as found out in the previous step.
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^2} = \dfrac{1}{2}{e^{i\dfrac{\pi }{2}}}\]
Now consider \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^2}}}\],
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{{2^2}}} = {\left( {\dfrac{1}{2}{e^{i\dfrac{\pi }{2}}}} \right)^2}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{1}{4}{e^{i\pi }}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{1}{{{2^2}}}{e^{i\pi }}\]
Next considering \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^3}}}\],
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{{2^3}}} = {\left( {\dfrac{{1 + i}}{2}} \right)^8}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = {\left( {\dfrac{1}{{\sqrt 2 }}{e^{i\dfrac{\pi }{4}}}} \right)^8}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{4^2}}}{e^{i2\pi }}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{2^4}}}{e^{i2\pi }}\]
Next considering \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^4}}}\],
$ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{{2^4}}} = {\left( {\dfrac{{1 + i}}{2}} \right)^{16}}$
$ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = {\left( {\dfrac{1}{{\sqrt 2 }}{e^{i\dfrac{\pi }{4}}}} \right)^{16}}$
$ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = \dfrac{1}{{{2^8}}}{e^{i4\pi }}$
There is a pattern following here :
$ \Rightarrow \dfrac{{1 + i}}{2} = \dfrac{1}{{\sqrt 2 }}{e^{i\dfrac{\pi }{4}}}$
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^2} = \dfrac{1}{2}{e^{i\dfrac{\pi }{2}}}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{1}{{{2^2}}}{e^{i\pi }}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{2^4}}}{e^{i2\pi }}\]
$ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = \dfrac{1}{{{2^8}}}{e^{i4\pi }}$
Now further simplifying the above expressions:
First consider \[{\left( {\dfrac{{1 + i}}{2}} \right)^2}\],
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^2} = \dfrac{1}{2}{e^{i\dfrac{\pi }{2}}}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^2} = \dfrac{1}{2}\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\]
\[\therefore {\left( {\dfrac{{1 + i}}{2}} \right)^2} = \dfrac{i}{2}\]
Now consider \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^2}}}\],
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{1}{{{2^2}}}{e^{i\pi }}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{1}{{{2^2}}}\left( {\cos \pi + i\sin \pi } \right)\]
\[\therefore {\left( {\dfrac{{1 + i}}{2}} \right)^4} = \dfrac{{ - 1}}{{{2^2}}}\]
Now consider \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^3}}}\],
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{2^4}}}{e^{i2\pi }}\]
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{2^4}}}\left( {\cos 2\pi + i\sin 2\pi } \right)\]
\[\therefore {\left( {\dfrac{{1 + i}}{2}} \right)^8} = \dfrac{1}{{{2^4}}}\]
Now consider \[{\left( {\dfrac{{1 + i}}{2}} \right)^{{2^4}}}\],
$ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = \dfrac{1}{{{2^8}}}{e^{i4\pi }}$
\[ \Rightarrow {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = \dfrac{1}{{{2^8}}}\left( {\cos 4\pi + i\sin 4\pi } \right)\]
\[\therefore {\left( {\dfrac{{1 + i}}{2}} \right)^{16}} = \dfrac{1}{{{2^8}}}\]
$\therefore $There is a pattern which is followed here which is if the power of \[\left( {\dfrac{{1 + i}}{2}} \right)\] is n, then the RHS of it has a denominator of 2 which will be having a power of ${2^{n - 1}}$. By applying the pattern it is :
\[\therefore {\left( {\dfrac{{1 + i}}{2}} \right)^{{2^n}}} = \dfrac{1}{{{2^{{2^{n - 1}}}}}}\]
Now substitute all the obtained expressions in the product expression:
$ \Rightarrow \left[ {1 + \left( {\dfrac{{1 + i}}{2}} \right)} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^2}} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^2}}}} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^3}}}} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^4}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^n}}}} \right]$
$ \Rightarrow \left[ {\dfrac{{3 + i}}{2}} \right]\left[ {1 + \dfrac{i}{2}} \right]\left[ {1 + \dfrac{{ - 1}}{4}} \right]\left[ {1 + \dfrac{1}{{{2^4}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
$ \Rightarrow \left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{4}} \right]\left[ {1 + \dfrac{1}{{{2^4}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
On further simplifying the 3rd term can be written as $\left[ {1 - \dfrac{1}{{{2^2}}}} \right]$, as there is another pattern that fits for solving the product which is:
\[ \Rightarrow \left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{2^2}}}} \right]\left[ {1 + \dfrac{1}{{{2^4}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]\]
Multiply and divide the expression with $\left[ {1 + \dfrac{1}{4}} \right]$ :
$ \Rightarrow \dfrac{1}{{\left[ {1 + \dfrac{1}{4}} \right]}}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 + \dfrac{1}{4}} \right]\left[ {1 - \dfrac{1}{4}} \right]\left[ {1 + \dfrac{1}{{{2^4}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
Here it is $\left[ {1 + \dfrac{1}{4}} \right]\left[ {1 - \dfrac{1}{4}} \right]$, which is in the form of $(a + b)(a - b),$which is equal to $({a^2} - {b^2})$
$ \Rightarrow \dfrac{1}{{\left[ {1 + \dfrac{1}{4}} \right]}}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{4^2}}}} \right]\left[ {1 + \dfrac{1}{{{2^4}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
$ \Rightarrow \dfrac{1}{{\left[ {\dfrac{5}{4}} \right]}}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{4^2}}}} \right]\left[ {1 + \dfrac{1}{{{4^2}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
Now $\left[ {1 - \dfrac{1}{{{4^2}}}} \right]\left[ {1 + \dfrac{1}{{{4^2}}}} \right]$ are in the form of $(a + b)(a - b),$which is equal to $({a^2} - {b^2})$
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{{16}^2}}}} \right]\left[ {1 + \dfrac{1}{{{2^8}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{{16}^2}}}} \right]\left[ {1 + \dfrac{1}{{{{16}^2}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
Now again $\left[ {1 - \dfrac{1}{{{{16}^2}}}} \right]\left[ {1 + \dfrac{1}{{{{16}^2}}}} \right]$ are in the form of $(a + b)(a - b),$which is equal to $({a^2} - {b^2})$
$\therefore $This process goes on till $\left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$ gets $\left[ {1 - \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
At the end the expression becomes:
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]\left[ {1 + \dfrac{1}{{{2^{{2^{n - 1}}}}}}} \right]$
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{2^{{2^n}}}}}} \right]$
Now simplifying the product:
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{3 + i}}{2}} \right]\left[ {\dfrac{{2 + i}}{2}} \right]\left[ {1 - \dfrac{1}{{{2^{{2^n}}}}}} \right]$
$ \Rightarrow \dfrac{4}{5}\left[ {\dfrac{{5 + 5i}}{4}} \right]\left[ {1 - \dfrac{1}{{{2^{{2^n}}}}}} \right]$
$ \Rightarrow (1 + i)\left[ {1 - \dfrac{1}{{{2^{{2^n}}}}}} \right]$
$\therefore \left[ {1 + \left( {\dfrac{{1 + i}}{2}} \right)} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^2}} \right]\left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^2}}}} \right] \cdot \cdot \cdot \cdot \cdot \left[ {1 + {{\left( {\dfrac{{1 + i}}{2}} \right)}^{{2^n}}}} \right] = (1 + i)\left[ {1 - \dfrac{1}{{{2^{{2^n}}}}}} \right]$

Note: Here while solving have to carefully observe through the question again and again and solve to check whether it is following a certain pattern, and apply the pattern to solve the rest of the problem.