Find limlimits_(n→∞)(dfrac4+3i5)^n
Find $\ \lim\limits_{n\to\infty}\left(\dfrac{4+3i}5\right)^n$.
Explanation of Method
Write $r=\frac{4+3i}5$ in polar form. Since $|r|=1$ exactly (a $3$–$4$–$5$ triangle), the only way $r^n$ could converge is if $r=1$ itself; otherwise $r^n=e^{in\theta}$ stays on the unit circle forever, and convergence hinges on whether $\theta/2\pi$ is rational.
Step 1 · Compute $|r|$
$$ |r|=\frac{|4+3i|}5=\frac{\sqrt{16+9}}5=\frac{\sqrt{25}}5=1 $$
Step 2 · Write $r=e^{i\theta}$
$$ r=\frac45+i\frac35, \qquad \theta=\arctan\frac34\approx0.6435\text{ rad}\approx36.87^\circ $$
Step 3 · Why $r^n$ cannot converge
If $r^n\to L$, then $|L|=\lim|r|^n=1$ and also $r^{n+1}/r^n=r\to1$ would force $r=1$ (taking the ratio of consecutive terms of a convergent geometric sequence). Since $r=\frac{4+3i}5\neq1$, the sequence $r^n$ cannot converge to any finite value — it must keep moving around the unit circle.
Step 4 · $\theta/\pi$ is irrational (Niven's theorem)
Niven's theorem classifies all angles $\theta$ with $\theta/\pi$ rational and $\cos\theta$ rational: the only such values are $\cos\theta\in\left\{0,\pm\frac12,\pm1\right\}$. Here $\cos\theta=\frac45$, which is rational but not in that finite exceptional list — so $\theta/\pi$ (and hence $\theta/2\pi$) must be irrational.
Step 5 · Consequence
Since $\theta/2\pi$ is irrational, Weyl's equidistribution theorem says $\{e^{in\theta}\}$ is dense (in fact equidistributed) on the unit circle — the sequence visits every neighborhood of every point on $|z|=1$ infinitely often, so it has no limit.
Verification · first few terms
$$ r^1=0.8+0.6i, \qquad r^2=\frac{(4+3i)^2}{25}=\frac{7+24i}{25}=0.28+0.96i, \qquad r^3=\frac{(7+24i)(4+3i)}{125}=\frac{-44+117i}{125}=-0.352+0.936i $$
All three have modulus exactly $1$, yet visibly different arguments — no sign of settling toward a limit.
Result
The accumulation set of this sequence is the entire unit circle — a case where $|r|=1$ but $r$ is not a root of unity, so the powers never repeat and never settle.