# Cauchy–Schwarz inequality for complex numbers

How can I prove the Cauchy– Schwarz inequality for two complex numbers? $$z_1=x_1+iy_1$$ $$z_2=x_2+iy_2$$

I can prove the triangle inequality for two complex numbers: $$|z_1+z_2|\le |z_1|+|z_2|.$$ But I cannot prove the Cauchy–Schwarz inequality: $$|z_1\cdot z_2|^2\le |z_2|^2|z_2|^2.$$

In my calculations, I always find the two expressions to be equal.

$a_1, a_2, \ldots, a_n \in \mathbb{C}$ and $b_1, b_2, \ldots, b_n \in \mathbb{C}$: when the $n=1$ $$|\sum_{j=1}^1 a_j \overline{b_j}|^2 \leq \sum_{j=1}^1 |a_j|^2 \sum_{j=1}^1 |b_j|^2$$

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use triangle inequality with on each component . – Theorem Jun 5 '12 at 15:23
@Mr Ares, what you want to prove? Maybe you mean the Cauchy-Schwarz for two complex numbers... – Hiperion Jun 5 '12 at 15:56
Clarify your question. – rschwieb Jun 5 '12 at 15:59
Should be $\mathrm{Re}(z_1\cdot\bar{z_2})^2\le|z_1|^2|z_2|^2$. – ziyuang Jun 5 '12 at 19:56
@Belgi: it is not strictly clear. If one treats $\mathbb{C}$ as a one complex dimensional complex vector space, then the standard inner product is indeed as you wrote $\langle z_1,z_2\rangle = z_1\bar{z_2}$. But there's also the possibility of viewing $\mathbb{C}$ as a two real dimensional vector space where the underlying field is $\mathbb{R}$. In this case the standard inner product on $\mathbb{R}^2$ can be written as $\langle z_1,z_2\rangle_{\mathbb{R}} = \mathrm{Re}(z_1\bar{z_2})$ as ziyuang wrote. (I fixed your typo, btw.) – Willie Wong Jun 6 '12 at 14:02

The Cauchy--Schwarz inequality is usually stated for vectors, not for just two numbers $z_1$ and $z_2$. In your case, if you consider numbers (i.e, the vectors of the inner product space $\mathbb C^1$), the Cauchy--Schwarz inequality is trivially true and indeed just equality: $$|z_1\bar{z}_2|=|z_1||z_2|.$$

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is there any difference between $|z_1\bar{z}_2|$ and $|z_1\cdot z_2|$ ? – user29646 Jun 6 '12 at 18:12
@MrAres The values $|z_1\bar{z}_2|$ and $|z_1\cdot z_2|$ are the same. However, $z_1\cdot z_2$ is not an inner product in $\mathbb C^1$, whereas $z_1\cdot\bar{z}_2$ is. – Artem Jun 6 '12 at 18:30

Every complex number $z$, can be written on the polar form:

$$z = re^{i \alpha}$$

Where $r$ and $\alpha$ are both real, and $r$ positive. The absolute value of this $z$ is $r$. Lets say we want to find $|z_{1}z_{2}|$. Rewrite in polar coordinates:

$$|z_{1}z_{2}| = |r_{1}e^{i \alpha_{1}}r_{2}e^{i \alpha_{2}}| = |r_{1}r_{2}e^{i ( \alpha_{1} + \alpha_{2})}|$$

And we see that our new absolute value is $r_{1}r_{2}$ since $\alpha_{2} + \alpha_{1}$ is still real (if it was complex then this would not be true). Which happens to be the same as $|z_{1}||z_{2}|$:

$$|z_{1}||z_{2}| = |r_{1}e^{i \alpha_{1}}||r_{2}e^{i \alpha_{2}}| = r_{1}r_{2}$$

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