Convergent Complex Sequence/Examples/(1 over 2 + i 4 over 5)^n

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Example of Convergent Complex Sequence

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

$z_n = \paren {\dfrac 1 2 + i \dfrac 4 5}^n$

Then:

$\displaystyle \lim_{n \mathop \to \infty} z_n = 0$


Proof

\(\ds \cmod {z_n}^2\) \(=\) \(\ds \cmod {\dfrac 1 2 + i \dfrac 4 5}^{2 n}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac 1 4 + \dfrac {16} {25} }^n\) Definition of Complex Modulus
\(\ds \) \(=\) \(\ds \paren {\dfrac {25 + 64} {100} }^n\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {89} {100} }^n\)
\(\ds \) \(\to\) \(\ds 0\) as $\cmod {\dfrac 2 3 + \dfrac {3 i} 4} < 1$

Thus $\cmod {z_n} \to 0$ and so $z_n \to 0$ as $n \to \infty$.

$\blacksquare$


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