Exists $\displaystyle\lim_{n\rightarrow \infty}\left(i + \left(\frac{4+3i}{5}\right)^n\right)$?

Question

Taking $\alpha = \frac{4+3i}{5}$ we have that $|\alpha|=1$, but on the other hand, we can say nothing about $1^n$ when $n \rightarrow \infty$.

Answer 1

The sequence $\left(\left(\frac{4+3i}5\right)^n\right)_{n\in\mathbb N}$ doesn't converge. If it did, let $l$ be its limit. Then\begin{align}l&=\lim_{n\to\infty}\left(\frac{4+3i}5\right)^{n+1}\\&=\lim_{n\to\infty}\left(\frac{4+3i}5\right)^n\left(\frac{4+3i}5\right)\\&=l\times\left(\frac{4+3i}5\right),\end{align}which is impossible, since this would mean that $l=0$ and the absolute value of each term of the sequence is $1$. So, your sequence also doesn't converge.

Answer 2

Note $\alpha$ has modulus $1$, and so represents a rotation. If you continually multiply it to itself, it simply rotates around the unit circle and never converges.

Answer 3

Note that $$(\frac {4+3i}{5})^n =\cos (n\theta)+i\sin (n\theta)$$ where $\cos (\theta)=4/5$ and $\sin (\theta )=3/5$.

Thus the sequence does not converge.

Answer 4

It is possible to expand this into lim n to infinity

i + summation k=0 to n of (n!*4^n * 3^k)/((n-k)!k! 4^k * 5^n ) * (cos(pi * k/2) + i * sin(pi * k/2) ) I cannot make any conclusion from this expansion. I hope it helps.