Limits of Real and Imaginary Parts

Theorem
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$.

Let $z_0 \in D$ be a complex number.

Suppose $f$ is continuous at $z_0$.

Then:


 * $(1): \quad \ds \lim_{z \mathop \to z_0} \map \Re {\map f z} = \map \Re {\lim_{z \mathop \to z_0} \map f z}$


 * $(2): \quad \ds \lim_{z \mathop \to z_0} \map \Im {\map f z} = \map \Im {\lim_{z \mathop \to z_0} \map f z}$

where:
 * $\map \Re {\map f z}$ denotes the real part of $\map f z$
 * $\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.

Proof
By definition of continuity:


 * $\forall \epsilon > 0: \exists \delta > 0: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$

Given $\epsilon > 0$, we find $\delta > 0$ so for all $z \in \C$ with $\cmod {z - z_0} < \delta$:

It follows that:


 * $\forall \epsilon > 0: \exists \delta > 0: \cmod {z - z_0} < \delta \implies \cmod {\map \Re {\map f z} - \map \Re {\map f {z_0} } } < \epsilon$

Then equation $(1)$ is proven by:

The proof for equation $(2)$ with imaginary parts follows when $\Re$ is replaced by $\Im$ in the equations above.