Multiplication of Real and Imaginary Parts

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Theorem

Let $w, z \in \C$ be complex numbers.

$(1)$ If $w$ is wholly real, then:

$\map \Re {w z} = w \, \map \Re z$

and:

$\map \Im {w z} = w \, \map \Im z$

$(2)$ If $w$ is wholly imaginary, then:

$\map \Re {w z} = -\map \Im w \, \map \Im z$

and:

$\map \Im {w z} = \map \Im w \, \map \Re z$

Here, $\map \Re z$ denotes the real part of $z$, and $\map \Im z$ denotes the imaginary part of $z$.

Proof

Assume that $w$ is wholly real.

Then:

\(\ds w z\)

\(=\)

\(\ds \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}\)

Definition of Complex Multiplication

\(\ds \)

\(=\)

\(\ds w \, \map \Re z + i w \, \map \Im z\)

as $\map \Re w = w$ and $\map \Im w = 0$

This equation shows that $\map \Re {w z} = w \, \map \Re z$, and $\map \Im {w z} = w \, \map \Im z$.

This proves $(1)$.

Now, assume that $w$ is wholly imaginary.

Then:

\(\ds w z\)

\(=\)

\(\ds \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}\)

\(\ds \)

\(=\)

\(\ds -\map \Im w \, \map \Im z) + i \, \map \Im w \, \map \Re z\)

as $\map \Re w = 0$

This equation shows that $\map \Re {w z} = -\map \Im w \, \map \Im z$, and $\map \Im {w c} = \map \Im w \, \map \Re z$.

This proves $(2)$.

$\blacksquare$