Convergent Complex Sequence/Examples/(cos pi over n + i sin pi over n)^2n+1

Example of Convergent Complex Sequence

Let $\sequence {z_n}$ be the complex sequence defined as:

$z_n = \paren {\cos \dfrac \pi n + i \sin \dfrac \pi n}^{2 n + 1}$

Then:

$\ds \lim_{n \mathop \to \infty} z_n = 1$

Proof

\(\ds z_n\)

\(=\)

\(\ds \paren {\cos \dfrac \pi n + i \sin \dfrac \pi n}^{2 n + 1}\)

\(\ds \)

\(=\)

\(\ds \cos \dfrac {\paren {2 n + 1} \pi} n + i \sin \dfrac {\paren {2 n + 1} \pi} n\)

De Moivre's Theorem

\(\ds \)

\(=\)

\(\ds \cos \paren {2 + \dfrac 1 n} \pi + i \sin \paren {2 + \dfrac 1 n} \pi\)

\(\ds \)

\(\to\)

\(\ds \cos 2 \pi + i \sin 2 \pi\)

as $\dfrac 1 n$ is a Basic Null Sequence

\(\ds \)

\(=\)

\(\ds 1\)

Cosine of Multiple of Pi, Sine of Multiple of Pi

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.2$. Sequences: Example $\text {(v)}$