I've been trying this for a while and discussed it with some friends but could not solve it. The exercise goes like this:

Let $(z_n)n\in \mathbb{N}$ the sequence of complex numbers defined by

$ z_1= 6i$

$z_{n+1}= \frac {3(1+i)z_n}{i}$

Prove that:

$ arg (z^2)= \begin{cases} \frac{3\pi}{2} \quad \text{if } n\equiv 0 \pmod 4\\ \pi \quad \text{if } n\equiv 1 \pmod 4\\ \frac{\pi}{2} \quad \text{if } n\equiv 2 \pmod 4\\ 0 \quad \text{if } n\equiv 3 \pmod 4\\ \end{cases}$

So, first I noticed that an induction would work. I tried to think about it in four different ones, or maybe as different sub-sequences of the original. In any case, I found the base cases and computed the first cases to have an idea of what is going on. Basically, my main issue is how to write the inductive step. I would write my attempts, but they are just basically isolated notes, drawings and not very clear strategies. I don`t really see how to connect my information with the form of the sequence and my inductive step, which has to do with the argument and the square of the argument. I do notice why this works $\pmod 4$ and what kind of graphic it draws, but that's it.

I think I'd do better with some help.

Thanks a lot

  • 1
    $\begingroup$ I presume you mean $z_{n+1}$, not $z_n+1$? $\endgroup$ – Lord Shark the Unknown Jun 22 '17 at 4:17
  • $\begingroup$ Thanks. I'll edit it. $\endgroup$ – Esteban Sargiotto Jun 22 '17 at 4:18
  • $\begingroup$ I presume you mean $\arg(z_n^2)$ not $\arg(z^2)$. $\endgroup$ – Lord Shark the Unknown Jun 22 '17 at 4:19
  • $\begingroup$ Well, what you say makes sense, but that's what the exercise says. $\endgroup$ – Esteban Sargiotto Jun 22 '17 at 4:24


$$\frac{3(1+i)}{i}=3(1-i)=3\sqrt 2 e^{-i\pi/4}$$

Hence, $\arg(z^2_{n+1})=\arg(z^2_n)-\pi/2$.


Write $w_n=z_n^2$. Then $w_1=-36$ and $$w_{n+1}=\left(\frac{3(1+i)}{i}\right)^2w_n =\frac{18 i}{-1}w_n=-18i w_n.$$ You should be able to derive a nice general formula for $w_n$.


If we write the sequence as

$$z_{n+1}= \frac {3(1+i)z_n}{i}= 3(1-i)z_n=Az_n,\quad z_0=6i$$

and ignore for the moment that $A$ is complex, then we can readily show by induction that

$$z_n=A^n z_0$$

Then we have

$$ z_n^2=A^{2n}z_0^2\\ \arg (z_n^2)=\tan^{-1}\left(\frac{\Im\{(3(1-i))^{2n}z_0^2\}}{\Re\{(3(1-i))^{2n}z_0^2\}} \right)=\tan^{-1}\left(\frac{\Im\{(1-i)^{2n}z_0^2\}}{\Re(1-i)^{2n}z_0^2\}} \right) $$

However, my results for the argument differ from yours and repeat in mod 4 as follows, starting at $n=0$,

$$\arg(z_{n=0...3})=[\pi,~\pi/2,~0,-\pi/2] $$

I have verified both my sequence and argument results numerically. Notice how this solution dovetails with @MarkViola.


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