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Toolkit 113

Complex Conjugates and Conjugate Roots

Complex Conjugates

z=a+bi⟹z‾=a−biz=a+bi\Longrightarrow \overline{z}=a-bi
z1+z2‾=z1‾+z2‾\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}
z1−z2‾=z1‾−z2‾\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}
z1z2‾=z1‾ z2‾\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}
zn‾=z‾,nn∈Z+\overline{z^n}=\overline{z}^{,n}\qquad n\in\mathbb{Z}^+
z=z‾⟺z∈Rz=\overline{z}\Longleftrightarrow z\in\mathbb{R}

Complex Conjugate Roots

Suppose P(x)∈R[x]P(x)\in\mathbb{R}[x] (P(x) is a polynomial with real roots).

z is a root of P(x)⟹z‾ is a root of P(x) tooz\text{ is a root of }P(x)\Longrightarrow \overline{z}\text{ is a root of }P(x)\text{ too}

Suppose P(x)∈Q[x]P(x)\in\mathbb{Q}[x] (P(x) is a polynomial with rational roots).

a+b5 is a root of P(x)⟹a−b5 is a root of P(x) tooa+b\sqrt5\text{ is a root of }P(x)\Longrightarrow a-b\sqrt5\text{ is a root of }P(x)\text{ too}

Instead of 5\sqrt5, we can consider 53\sqrt[3]{5} or any other irrational number.

Proof

Proof of Complex Conjugate Properties

z1+z2‾=z1‾+z2‾\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}
z1−z2‾=z1‾−z2‾\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}
z1z2‾=z1‾ z2‾\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}

Let

z1=a+bi,z2=c+di.z_1=a+bi,\qquad z_2=c+di.
z1+z2‾=a+bi+c+di‾=a+c+(b+d)i‾=a+c−(b+d)i=(a−bi)+(c−di)=z1‾+z2‾.\begin{aligned}\overline{z_1+z_2}&=\overline{a+bi+c+di}\\&=\overline{a+c+(b+d)i}\\&=a+c-(b+d)i\\&=(a-bi)+(c-di)\\&=\overline{z_1}+\overline{z_2}.\end{aligned}
z1−z2‾=a+bi−c−di‾=a−c+(b−d)i‾=a−c−(b−d)i=(a−bi)−(c−di)=z1‾−z2‾.\begin{aligned}\overline{z_1-z_2}&=\overline{a+bi-c-di}\\&=\overline{a-c+(b-d)i}\\&=a-c-(b-d)i\\&=(a-bi)-(c-di)\\&=\overline{z_1}-\overline{z_2}.\end{aligned}
z1z2‾=(a+bi)(c+di)‾=(ac−bd)+(ad+bc)i‾=(ac−bd)−(ad+bc)i=(a−bi)(c−di)=z1‾ z2‾.\begin{aligned}\overline{z_1z_2}&=\overline{(a+bi)(c+di)}\\&=\overline{(ac-bd)+(ad+bc)i}\\&=(ac-bd)-(ad+bc)i\\&=(a-bi)(c-di)\\&=\overline{z_1}\,\overline{z_2}.\end{aligned}
zn‾=z‾ n,n∈Z+\overline{z^n}=\overline{z}^{\,n},\qquad n\in\mathbb{Z}^{+}

By the 3rd property,

z2‾=z⋅z‾=z‾⋅z‾=z‾,2.\overline{z^2}=\overline{z\cdot z}=\overline{z}\cdot\overline{z}=\overline{z}^{,2}.
z3‾=z2⋅z‾=z‾,2⋅z‾=z‾,3.\overline{z^3}=\overline{z^2\cdot z}=\overline{z}^{,2}\cdot\overline{z}=\overline{z}^{,3}.
⋮\vdots
zn‾=z‾ n.\overline{z^n}=\overline{z}^{\,n}.

We can prove rigorously by induction.

z=z‾  ⟺  z∈R.z=\overline{z}\iff z\in\mathbb{R}.

Let

z=a+bi.z=a+bi.
z=z‾  ⟺  a+bi=a−bi  ⟺  2bi=0  ⟺  b=0  ⟺  z∈R.z=\overline{z}\iff a+bi=a-bi\iff 2bi=0\iff b=0\iff z\in\mathbb{R}.
anzn+an−1zn−1+⋯+a1z+a0‾=0\overline{a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0}=0
anz‾ n+an−1z‾ n−1+⋯+a1z‾+a0=0a_n\overline{z}^{\,n}+a_{n-1}\overline{z}^{\,n-1}+\cdots+a_1\overline{z}+a_0=0
P(z‾)=0P(\overline{z})=0

Therefore, if z is a root of a polynomial with real coefficients, then its complex conjugate is also a root.

Proof for Rational Coefficients

Suppose P(x)∈Q[x]P(x)\in\mathbb{Q}[x] and a+b5a+b\sqrt{5} is a root.

P(a+b5)=0P(a+b\sqrt{5})=0

The algebraic conjugate of a+b5a+b\sqrt{5} is

a−b5a-b\sqrt{5}

Because the coefficients of P(x) are rational, the conjugate property gives

P(a−b5)=0P(a-b\sqrt{5})=0

Therefore, roots involving square roots occur in conjugate pairs.

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