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$$\lim\limits_{n \rightarrow \infty} \left(\frac{i}{1+i}\right)^n$$

I think the limit is $0$; is it true that $\forall a,b\in \Bbb C$, if $|a|<|b|$ then $\lim\limits_{n\rightarrow \infty}\left(\frac{a}{b}\right)^n=0$?

I would like to see a proof, if possible. Thank you

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3 Answers 3

up vote 10 down vote accepted

You are right. For a proof, just observe that if $\frac{|a|}{|b|}<1$, then $$ \left\|\left(\frac ab\right)^n\right\| = \left(\frac{|a|}{|b|}\right)^n \to 0 $$ as $n\to\infty$.

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$\frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i(1-i)}{2}=\dfrac{e^{\frac{\pi}{2}i}e^{-\frac{1}{4}\pi i}}{\sqrt{2}} = \dfrac{e^{\frac{\pi}{4}i}}{\sqrt{2}}$, so $(\frac{i}{1+i})^n = \dfrac{e^{\frac{n\pi}{4}i}}{(\sqrt{2})^n}$

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Observe that $$\frac{i}{1+i} =\frac{1}{2}+\frac{i}{2}$$ Now define $$F_n=\Big(\frac{i}{1+i}\Big)^n$$ Then $$F_1=\frac{1}{2}+\frac{i}{2}$$ $$F_2=\frac{i}{2}$$ $$F_3=-\frac{1}{4}+\frac{i}{4}$$ $$F_4=-\frac{1}{4}$$ $$F_5=-\frac{1}{8}-\frac{i}{8}$$

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its interesting that all the 4th powers in the limit are real numbers, clearly going to 0 by pre-cal principles. So you can at least say that if the limit exists, then it must be 0, even with no more than that observation. –  TCC Aug 15 at 14:37
    
Observation is totally unsufficient. I just gave that for illustration purposes. The answers are really good so I just added a small blabla ! Cheers :-) –  Claude Leibovici Aug 15 at 14:41
    
yes it is insufficient for a complete solution to this problem ... still i looked at your blabla and thought it was neat that there is a subsequence of real numbers that obviously go to 0 ... there are probably other sequences of complex numbers that have real number subsequences with obvious limits ... that would be sometimes useful, if for nothing else than to preclude the possibility of a non-real limit to the sequence –  TCC Aug 15 at 14:51
    
@TCC: Are you surprised that all the 4th powers in the series are real numbers? Have you tried visualizing the problem in the complex coordinate plane? $\frac{i}{1+i} = \frac{-1+i}{2}$, which is a vector $45^\circ$ away from the real axis ($135^\circ$ away from the positive side of the real axis), so every 4th power is at a multiple of $4\times45^\circ$, i.e., a multiple of $180^\circ$, and therefore is real. –  Scott Aug 15 at 15:09
    
@Scott, no I'm not surprised, but i agree with your explanation ... more generally we can see by the same reasoning that any complex number has this property if and only if its argument (angle to positive real axis) is a rational multiple of pi... in this case (3/4)pi... the period is always the denominator of the reduced form of the ratio –  TCC Aug 15 at 16:27

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