For this question I am not getting any start .
Can somebody provide me a hint
Prove the product of complex numbers
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$\begingroup$
complex-numbers
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1Have you tried induction on $n$? – 2017-02-01
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1also you could write $1+i=\sqrt{2}e^{i\pi /4}$ – 2017-02-01
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0@AlexMathers I don't want to use induction – 2017-02-01
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0@Basti then also itis difficult to solve – 2017-02-01
1 Answers
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Let $\omega=\dfrac{1+i}{2}$ \begin{align*} \omega^2 &= \frac{1+2i-1}{4} \\ &= \frac{i}{2} \\ 1+\omega &= \frac{1-\omega^2}{1-\omega} \\ (1+\omega)(1+\omega^2) &= \frac{1-\omega^4}{1-\omega} \\ (1+\omega)(1+\omega^2) \ldots (1+\omega^{2^n}) &= \frac{1-\omega^{2^{n+1}}}{1-\omega} \\ &= \left[1-\left( \frac{i}{2} \right)^{2^n} \right] \times \frac{1}{1-\omega} \\ &= \left[ 1-\frac{(-1)^{2^{n-1}}}{2^{2^n}} \right] \times \frac{2}{1-i} \\ &= \left( 1-\frac{1}{2^{2^n}}\right)\times \frac{2(1+i)}{1-i^2} \tag{$n\ge 2$} \\ &= \left( 1-\frac{1}{2^{2^n}}\right)(1+i) \end{align*}