0
$\begingroup$

How to evaluate the following limit? $$ \lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n $$ Here $i=\sqrt{-1}$.

I got: $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n = \lim_{n\to \infty} \frac{(i-1)^n}{n(\sqrt{2})^n} $$ I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?

$\endgroup$
  • 1
    $\begingroup$ It's very hard to understand what you wrote. Try mathjax $\endgroup$ – DonAntonio Nov 24 '18 at 21:38
  • $\begingroup$ Express $(1+i)/\sqrt{2}=\exp(i\pi/4)$. $\endgroup$ – Diger Nov 24 '18 at 21:39
  • 1
    $\begingroup$ L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $\sqrt{blah blah}$. A limit of a function f(x) from $n \to \infty$ can be written $\lim_{n\to \infty} f(x)$. Just a few pointers. $\endgroup$ – Namaste Nov 24 '18 at 21:43
  • 1
    $\begingroup$ If your function $f(x)$ is a fraction, you can write it as follows $f(x) = \frac{"numerator here"}{"denominator here"}$ $\endgroup$ – Namaste Nov 24 '18 at 21:46
  • $\begingroup$ Write $\frac{1+i}{\sqrt{2}}=e^{i\pi/4}\implies (\frac{1+i}{\sqrt{2}})^n=e^{i\ n\pi/4}$ $\endgroup$ – Shubham Johri Nov 24 '18 at 21:57
4
$\begingroup$

We have $$ \left|\frac{1+i}{\sqrt{2}}\right|=1 $$ Thus $$ \left|\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^{\!n}\right|=\frac{1}{n} $$

$\endgroup$
  • $\begingroup$ egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2 $\endgroup$ – KarmaL Nov 25 '18 at 9:25
  • $\begingroup$ @L.Spy $\sqrt{(1/\sqrt{2})^2+(1/\sqrt{2})^2}=\sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$. $\endgroup$ – egreg Nov 25 '18 at 9:40
1
$\begingroup$

$$\lim_{n\to \infty} |\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n|=\lim_{n\to \infty} {1\over n}=0$$therefore $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.