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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 $n=1$ $$|\sum_{j=1}^1 a_j \overline{b_j}|^2 \leq \sum_{j=1}^1 |a_j|^2 \sum_{j=1}^1 |b_j|^2$$

grantfgates
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    use triangle inequality with on each component . – Theorem Jun 05 '12 at 15:23
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    @Mr Ares, what you want to prove? Maybe you mean the Cauchy-Schwarz for two complex numbers... – Hiperion Jun 05 '12 at 15:56
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    Clarify your question. – rschwieb Jun 05 '12 at 15:59
  • Should be $\mathrm{Re}(z_1\cdot\bar{z_2})^2\le|z_1|^2|z_2|^2$. – Ziyuan Jun 05 '12 at 19:56
  • @ziyuang Dear Sir this is inequality but the point is that, this is not really Cauchy–Schwarz inequality –  Jun 05 '12 at 20:05
  • @MrAres , this is CS inequality. CS inequality is about inner product, not product. – Ziyuan Jun 05 '12 at 20:08
  • @ziyuang so what is this one!?

    proving-the-schwarz-inequality-for-complex-numbers-using-induction

     –  Jun 05 '12 at 23:49
  • I think the confusion here is because the CS ineuality talks about inner products - but please note that the standard inner product on $\mathbb{C}$ is $\langle z_1,z_2 \rangle = z_1\bar{z_2}$. – Belgi Jun 06 '12 at 13:36
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    @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 06 '12 at 14:02

2 Answers2

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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: $$ |z_1\bar{z}_2|=|z_1||z_2|. $$

Artem
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  • is there any difference between $|z_1\bar{z}_2|$ and $|z_1\cdot z_2|$ ? –  Jun 06 '12 at 18:12
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    @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 06 '12 at 18:30
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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 $|z_{1}||z_{2}|$:

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

Eirik Hovind
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