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Formulas/maths/M2 Modulus Argument/Properties of Conjugate

Properties of Conjugate

Four fundamental identities. z·z̄ = a²+b² = |z|² is the most used — it converts division into multiplication. Also: conjugate distributes over addition and multiplication: z₁+z₂ bar = z̄₁+z̄₂, and z₁z₂ bar = z̄₁·z̄₂.
Derivation

Let z=a+ibz = a + ib and zˉ=aib\bar{z} = a - ib. All four identities follow by direct computation from this definition.

Identity 1 — Double Conjugate

zˉˉ=(aib)=a+ib=z\bar{\bar{z}} = \overline{(a - ib)} = a + ib = z

Taking the conjugate twice returns the original number. The conjugate is an involution.

Identity 2 — Sum

z+zˉ=(a+ib)+(aib)=2a=2Re(z)z + \bar{z} = (a + ib) + (a - ib) = 2a = 2\operatorname{Re}(z)

This is the most-used form for extracting the real part of a complex number. Rearranged: Re(z)=z+zˉ2\operatorname{Re}(z) = \dfrac{z + \bar{z}}{2}.

Identity 3 — Difference

zzˉ=(a+ib)(aib)=2ib=2iIm(z)z - \bar{z} = (a + ib) - (a - ib) = 2ib = 2i\operatorname{Im}(z)

Rearranged: Im(z)=zzˉ2i\operatorname{Im}(z) = \dfrac{z - \bar{z}}{2i}. Note the denominator is 2i2i, not 22.

Identity 4 — Product

zzˉ=(a+ib)(aib)=a2+b2=z2z\bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2

This is the most important identity in the chapter. It converts the product of two conjugates into a real number — which is why multiplying numerator and denominator by the conjugate clears the denominator in division.

Distributive Properties

Over addition:

z1+z2=(a1+a2)+i(b1+b2)=(a1+a2)i(b1+b2)=zˉ1+zˉ2\overline{z_1 + z_2} = \overline{(a_1 + a_2) + i(b_1 + b_2)} = (a_1 + a_2) - i(b_1 + b_2) = \bar{z}_1 + \bar{z}_2

Over multiplication:

z1z2=(a1a2b1b2)+i(a1b2+a2b1)=(a1a2b1b2)i(a1b2+a2b1)\overline{z_1 z_2} = \overline{(a_1 a_2 - b_1 b_2) + i(a_1 b_2 + a_2 b_1)} = (a_1 a_2 - b_1 b_2) - i(a_1 b_2 + a_2 b_1) zˉ1zˉ2=(a1ib1)(a2ib2)=(a1a2b1b2)i(a1b2+a2b1)\bar{z}_1 \bar{z}_2 = (a_1 - ib_1)(a_2 - ib_2) = (a_1 a_2 - b_1 b_2) - i(a_1 b_2 + a_2 b_1)

Both expressions are identical, so z1z2=zˉ1zˉ2\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2.

By induction this extends to any finite product: z1z2zn=zˉ1zˉ2zˉn\overline{z_1 z_2 \cdots z_n} = \bar{z}_1 \bar{z}_2 \cdots \bar{z}_n, and in particular zn=zˉn\overline{z^n} = \bar{z}^n.