The complex field

A complex number is an ordered pair $(a,b)$ of real numbers. Equality, addition, and multiplication are defined as follows:

• $(a,b)$ and $(c,d)$ are equal iff $a=c$ and $b=d$.
• $(a,b)+(c,d)=(a+c,b+d)$.
• $(a,b)(c,d)=(ac-bd,ad+bc)$.

These definitions turn the set of all complex numbers into a field.

If $z$ and $w$ are complex numbers, then

• $\overline{z+w}=\overline{z}+\overline{w}$
• $\overline{zw}=\overline{z}\cdot \overline{w}$
• $z+\overline{z}=2\mathrm{Re}(z)$
• $z-\overline{z}=2i\mathrm{Im}(z)$
• $z\overline{z}$ is real and positive except when $z=0$.

The Cauchy-Schwarz inequality in complex numbers

Recall the CS inequality in $\mathbb{R}$:

$${\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2\le \sum _{i=1}^n{a}_i^2\sum _{i=1}^n{b}_i^2$$

This can be interpreted as saying the dot product of two vectors in ${\mathbb{R}}^n$ is less than or equal to the product of their magnitudes:

$$|\mathbf{v}\cdot \mathbf{w}|\le \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert$$

If we apply the same definition to vectors over $\mathbb{C}$, we get the CS inequality in complex numbers.

$${\left|\sum _{j=1}^n{a}_j{\overline{b}}_j\right|}^2\le \sum _{j=1}^n{\left|a_j\right|}^2\sum _{j=1}^n{\left|b_j\right|}^2.$$

Proof on Rudin, p14.

Neat proof by Mohak:
From the CS inequality for real numbers and the triangle inequality for complex numbers, we have

$$\underset{\text{triangle inequality}}{\underbrace{{\left|\sum _{i=1}^n{a}_i\overline{b_i}\right|}^2\le {\left(\sum _{i=1}^n|a_i\overline{b_i}|\right)}^2}}=\underset{\text{CS inequality for reals}}{\underbrace{{\left(\sum _{i=1}^n|a_i||b_i|\right)}^2\le \sum _{i=1}^n{\left|a_i\right|}^2\sum _{i=1}^n{\left|b_i\right|}^2}}.$$