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Formulas/maths/M2 Modulus Argument/Properties of Modulus

Properties of Modulus

Modulus is multiplicative — it converts products into products of real numbers. Also: |Re(z)| ≤ |z| and |Im(z)| ≤ |z|, since each component cannot exceed the full distance from origin.
Derivation

The key to all modulus properties is the identity zzˉ=∣z∣2z\bar{z} = |z|^2. Once this is established, every property follows from algebra.

Property 1 — Modulus of a Product

Claim: ∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1||z_2|

Proof:

∣z1z2∣2=(z1z2)(z1z2)‾=(z1z2)(zˉ1zˉ2)=(z1zˉ1)(z2zˉ2)=∣z1∣2∣z2∣2|z_1 z_2|^2 = (z_1 z_2)\overline{(z_1 z_2)} = (z_1 z_2)(\bar{z}_1 \bar{z}_2) = (z_1 \bar{z}_1)(z_2 \bar{z}_2) = |z_1|^2 |z_2|^2

Taking square roots (both sides non-negative):

∣z1z2∣=∣z1∣∣z2∣■|z_1 z_2| = |z_1||z_2| \qquad \blacksquare

Property 2 — Modulus of a Quotient

Claim: ∣z1z2∣=∣z1∣∣z2∣\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}, z2≠0z_2 \neq 0

Apply Property 1 to z1=z1z2⋅z2z_1 = \dfrac{z_1}{z_2} \cdot z_2:

∣z1∣=∣z1z2∣⋅∣z2∣  ⟹  ∣z1z2∣=∣z1∣∣z2∣■|z_1| = \left|\frac{z_1}{z_2}\right| \cdot |z_2| \implies \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad \blacksquare

Property 3 — Modulus of Conjugate

Claim: ∣zˉ∣=∣z∣|\bar{z}| = |z|

∣zˉ∣2=zˉ zˉ‾=zˉ⋅z=zzˉ=∣z∣2|\bar{z}|^2 = \bar{z}\,\overline{\bar{z}} = \bar{z} \cdot z = z\bar{z} = |z|^2

So ∣zˉ∣=∣z∣|\bar{z}| = |z|. Geometrically: conjugation reflects across the real axis, preserving the distance from the origin. ■\blacksquare

Property 4 — Modulus of a Power

Claim: ∣zn∣=∣z∣n|z^n| = |z|^n

Apply Property 1 repeatedly:

∣zn∣=∣z⋅z⋯z∣=∣z∣∣z∣⋯∣z∣=∣z∣n■|z^n| = |z \cdot z \cdots z| = |z||z|\cdots|z| = |z|^n \qquad \blacksquare

Component Bounds

Since a2≤a2+b2a^2 \leq a^2 + b^2 and b2≤a2+b2b^2 \leq a^2 + b^2:

∣Re⁡(z)∣=∣a∣≤a2+b2=∣z∣|\operatorname{Re}(z)| = |a| \leq \sqrt{a^2 + b^2} = |z| ∣Im⁡(z)∣=∣b∣≤a2+b2=∣z∣|\operatorname{Im}(z)| = |b| \leq \sqrt{a^2 + b^2} = |z|

Each component is at most as large as the full modulus. This is used constantly in bounding arguments: to show ∣z∣|z| is small, it suffices to bound ∣Re⁡(z)∣|\operatorname{Re}(z)| and ∣Im⁡(z)∣|\operatorname{Im}(z)| separately.