Multiplication of Real and Imaginary Parts
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Two results -- or is it 4??
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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 $(2)$.
$\blacksquare$