Convergence of Complex Conjugate of Convergent Complex Sequence
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Theorem
Let $z \in \C$.
Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence converging to $z$.
Then:
- $\overline {z_n} \to \overline z$
Proof
Let $\epsilon > 0$.
Since $z_n \to z$, from the definition of convergence, we can find $N \in \N$ such that:
- $\cmod {z_n - z} < \epsilon$
From Complex Modulus equals Complex Modulus of Conjugate, we have:
- $\cmod {\overline {z_n - z} } = \cmod {z_n - z}$
From Difference of Complex Conjugates, we have:
- $\cmod {z_n - z} = \cmod {\overline {z_n} - \overline z}$
So we have:
- $\cmod {\overline {z_n} - \overline z} < \epsilon$
for each $n \ge N$.
Since $\epsilon$ was arbitrary, we have:
- $\overline {z_n} \to \overline z$
$\blacksquare$