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Complex Conjugates and Conjugate Roots

Lesson · Intermediate

Algebra: Complex Numbers

Complex Conjugates

z=a+bi⟹z‾=a−biz=a+bi\Longrightarrow \overline{z}=a-bi
z‾‾=z\overline{\overline{z}}=z
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}
zz‾=∣z∣2z\overline{z}=|z|^2
(z1z2)‾=z1‾z2‾z2≠0\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} \qquad z_2\ne0
zn‾=z‾ nn∈Z+\overline{z^n}=\overline{z}^{\,n} \qquad n\in\mathbb Z^+
z=z‾  ⟺  z∈Rz=\overline{z}\iff z\in\mathbb R

Complex Conjugate Roots

Suppose P(x) has real coefficients.

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) has rational coefficients.

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}
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