Modulus in Terms of Conjugate

Theorem

Let $z = a + i b$ be a complex number.

Let $\cmod z$ be the modulus of $z$.

Let $\overline z$ be the conjugate of $z$.

Then:

$\cmod z^2 = z \overline z$

Proof

Let $z = a + i b$.

Then:

\(\ds z \overline z\)

\(=\)

\(\ds a^2 + b^2\)

Product of Complex Number with Conjugate

\(\ds \)

\(=\)

\(\ds \cmod z^2\)

Definition of Complex Modulus

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1.2$. The Algebraic Theory: $(1.9)$
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups
• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms
• 1998: Yoav Peleg, Reuven Pnini and Elyahu Zaarur: Quantum Mechanics ... (previous) ... (next): Chapter $2$: Mathematical Background: $2.1$ The Complex Field $C$
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): conjugate (of a complex number)