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

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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.)2012-06-06

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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|. $

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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.2012-06-06
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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}$