Why does the numerator become 1 in the root test for solving series?

[DottZakapa]

Homework Statement

solving a complex series applying the root test

Relevant Equations

series with complex numbers

in order to solve a series, the root test is applied and I have this limit
$\lim_{n \rightarrow +\infty} \sqrt[n] {\left| {\frac {i^{\frac { n} {3}}} { \frac {2n} {3} +1}} \right| }$

I don't understand why at the second step the numerator becomes 1, cannot recall why it becomes 1, that is:

$\lim_{n \rightarrow +\infty} \sqrt[n] { {\frac {1} { \frac {2n} {3} +1}} }$

 

[fresh_42]

Is this supposed to be $\displaystyle{ i ^{n/3}}$ in the numerator?

 

[DottZakapa]

Is this supposed to be $\displaystyle{ i ^{n/3}}$ in the numerator?

yes

 

[Mark44]

Is this supposed to be $\displaystyle{ i ^{n/3}}$ in the numerator?

I've fixed it...

 

[Mark44]

Homework Statement:: solving a complex series applying the root test
Relevant Equations:: series with complex numbers

I don't understand why at the second step the numerator becomes 1, cannot recall why it becomes 1

The principal cube root of i (i.e., $i^{1/3}$) has a modulus of 1. Any integer power of this number produces a different angle, but the modulus remains 1.

 

[fresh_42]

yes

Then how do you calculate $\left| i^n\right|$?

 

[RPinPA]

Then how do you calculate $|i^n|$?

Write $i$ as $e^{i\pi/2}$. Then it's clear that $\left |i^n \right | = \left | e^{i n \pi/2} \right | = 1$ for any real n.

 

[fresh_42]

Write $i$ as $e^{i\pi/2}$. Then it's clear that $\left |i^n \right | = \left | e^{i n \pi/2} \right | = 1$ for any real n.

Would have been nice to read this from the OP.