$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$

Question

Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.

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I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.

Answer 1

Hints:

• $(1-a) \cdot (1+a) = 1 - a^2\,$, $\,(1-a) \cdot (1+a)(1+a^2) = 1 - a^4\,$, $\;\ldots$

• $(1+i)^2 = 2i$

• $1 - \dfrac{1+i}{2}=\dfrac{1-i}{2} = \dfrac{1}{1+i}$

Answer 2

Hint: Use the general case $$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})=\dfrac{(1-x^{2^{n+1}})}{1-x}$$

Answer 3

Hint: Use mathematical induction. Also, for $n > 2$, $$\left(\frac{1+i}{2}\right)^{2^{n}} = \frac{1}{2^{2^{n-1}}},$$ and $$\left(1-\frac{1}{2^{2^{n-1}}}\right)\left(1+\frac{1}{2^{2^{n-1}}}\right)=1-\frac{1}{2^{2^{n}}}.$$

Answer 4

Notice that, for any $x \neq 1,$ \begin{align*} \prod_{k=0}^n(1+x^{2^k})&=(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})\\&=\frac{(1-x)(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&=\frac{(1-x^2)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&\vdots\\&=\frac{1-x^{2^{n+1}}}{1-x}. \end{align*}

Thus, \begin{align*}\prod_{k=0}^n\left[1+\left(\frac{1+i}{2}\right)^{2^k}\right]&=\frac{1-\left(\dfrac{1+i}{2}\right)^{2^{n+1}}}{1-\dfrac{1+i}{2}}=\frac{1-\left[\left(\dfrac{1+i}{2}\right)^2\right]^{2^{n}}}{\dfrac{1}{1+i}}\\&=\left[1-\left(\frac{i}{2}\right)^{2^n}\right](1+i)\\&=\left(1-\frac{1}{2^{2^n}}\right)(1+i).\end{align*}