Let $z_n = \left(\frac{1+i}{3}\right)^n$ be a complex sequence. Show that $(z_n)$ converges.

I'm unsure how to do this because I've only just started learning about complex sequences. If this were real and, say we replaced $i$ with just $1$, I would note that $2^n < 3^n$ for all $n \geq 1$.

That said, I wonder if it's okay to use the absolute value of the complex number in the numerator and note that $(\sqrt{2})^n < 3^n$ to see that the sequence is convergent, or is this totally wrong?

  • 1
    $\begingroup$ Yes, you can note that $|(1 + i)/3| < 1$. Note that $|z_n| = |z_n - 0$, i.e. the distance from $0$. $\endgroup$
    – AJY
    Sep 21, 2016 at 0:20
  • $\begingroup$ @Ed generally, it's the absolute value of the complex number you are interested in. Convert the complex number into e-power format and see... $\endgroup$
    – imranfat
    Sep 21, 2016 at 1:27
  • $\begingroup$ @imranfat Do you mean that, when looking at the convergence of a complex sequence, you're more interested in the absolute value of the number than the number itself? I've read ahead a little and see the term "radius of convergence" banded around, so that seems to fit with that explanation. $\endgroup$ Sep 21, 2016 at 1:29
  • $\begingroup$ Consider a complex number of the form $r^n(cos\theta+i*sin\theta)^n$ and let $n$ go to infinity $\endgroup$
    – imranfat
    Sep 21, 2016 at 1:30
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    $\begingroup$ Yes indeed, but you also have to consider the situation $r=1$. Take for example $i$ and raise that to the power $n$. Any point on the unit circle will stay on the unit circle after subjected to powers. But that is not convergent either... $\endgroup$
    – imranfat
    Sep 21, 2016 at 2:10

1 Answer 1


Yes, the key here is that $$ \left| {1 + i \over 3} \right| \leq {2 \over 3} < 1. $$

  • $\begingroup$ Wonderful! Thanks a lot. $\endgroup$ Sep 21, 2016 at 0:22

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