Imaginary Part as Mapping is Endomorphism for Complex Addition

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Theorem

Let $\struct {\C, +}$ be the additive group of complex numbers.

Let $\struct {\R, +}$ be the additive group of real numbers.

Let $f: \C \to \R$ be the mapping from the complex numbers to the real numbers defined as:

$\forall z \in \C: \map f z = \map \Im z$

where $\map \Im z$ denotes the imaginary part of $z$.


Then $f: \struct {\C, +} \to \struct {\R, +}$ is a group epimorphism.


Its kernel is the set:

$\map \ker f = \R$

of (wholly) real numbers.


Proof

From Imaginary Part as Mapping is Surjection, $f$ is a surjection.


Let $z_1, z_2 \in \C$.

Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.

Then:

\(\ds \map f {z_1 + z_2}\) \(=\) \(\ds \map \Im {z_1 + z_2}\) Definition of $f$
\(\ds \) \(=\) \(\ds \map \Im {x_1 + i y_1 + x_2 + i y_2}\) Definition of $z_1$ and $z_2$
\(\ds \) \(=\) \(\ds y_1 + y_2\) Definition of Imaginary Part
\(\ds \) \(=\) \(\ds \map \Im {z_1} + \map \Im {z_2}\) Definition of Imaginary Part
\(\ds \) \(=\) \(\ds \map f {z_1} + \map f {z_2}\) Definition of $f$

So $f$ is a group homomorphism.

Thus $f$ is a surjective group homomorphism and therefore by definition a group epimorphism.


Finally:

$\forall y \in \R: \map \Im {x + 0 i} = 0 i = 0$

It follows from Complex Addition Identity is Zero that:

$\map \ker f = \set {x: x \in \R} = \R$

$\blacksquare$


Also see


Sources

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