Complex Number equals Negative of Conjugate iff Wholly Imaginary
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Theorem
Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Then $\overline z = -z$ if and only if $z$ is wholly imaginary.
Proof
Let $z = x + i y$.
Then:
\(\ds \overline z\) | \(=\) | \(\ds -z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x - i y\) | \(=\) | \(\ds -\left({x + i y}\right)\) | Definition of Complex Conjugate | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds +x\) | \(=\) | \(\ds -x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 0\) |
Hence by definition, $z$ is wholly imaginary.
$\Box$
Now suppose $z$ is wholly imaginary.
Then:
\(\ds \overline z\) | \(=\) | \(\ds 0 - i y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -i y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\left({0 + i y}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -z\) |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1.2$. The Algebraic Theory