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I see this all the time in Mathematica output as well as in text, such as near the top of the Wikipedia Beta function page.

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For a complex number $x \in \mathbb{C}$, you can write it as $x = a+bi$ where $a$ and $b$ are real numbers and $i$ is the imaginary number. $Re(x) = a$, it is referring to the "real part" of $x$. Similarly, there is a function called $Im$ such that $Im(x)=b$. – tomcuchta Oct 6 '11 at 2:48
Sometimes, you'll see $\Re z$ and $\Im z$ used instead of $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$. – J. M. Oct 6 '11 at 3:23
As a tiny Mathematica tip: whenever you see some function you don't quite understand in the output, highlight the name of the function (by double-clicking, for instance) and press the F1 key. – J. M. Oct 6 '11 at 3:26
up vote 13 down vote accepted

The real part of the complex number x. If you haven't seen complex numbers before, they're a two-dimensional version of the normal real numbers.

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If I may add (that is someone correct me if im wrong), the $\text{IM}(\cdot)$ part also is the real number however it is the on the imaginary axis in the complex plane. – Tyler Hilton Oct 6 '11 at 3:42
@Tyler: $\mathrm{Im}(z)$ is certainly real; what you probably had in mind is that if one considers the Argand plane, $\mathrm{Im}(z)$ is equivalent to the vertical coordinate of the point $z$. – J. M. Oct 6 '11 at 3:49

If a complex number $z$ is written as $z = a + bi$, then Re$(z) = a$ and Im$(z) = b$. (At risk of stating the obvious, "Re" stands for "Real" and "Im" stands for "Imaginary".)

If we visualize complex numbers as vectors in $\mathbb{R}^2$, Re is the projection onto the real axis, and Im is onto the imaginary axis. So $z = \mathrm{Re}(z) + \mathrm{Im}(z)i$.

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Do ALL real numbers get mapped on to the unit circle in the complex plane (or the Argand plane)? – Tyler Hilton Oct 6 '11 at 4:34
@Tyler: No, only those numbers with an absolute value of 1 can lie in the unit circle. – J. M. Oct 6 '11 at 7:39
@Tyler: The only real numbers on the unit circle of the complex plane are +1, -1, since real numbers are restricted to the real line and there are only these two points of intersection. There are an infinite number of complex numbers on the unit circle, of course. – Fixee Oct 6 '11 at 15:57

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