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.
What does $\mathrm{Re}(x)$ mean?
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$\begingroup$
notation
complex-numbers
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4For 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 $Im(x)=b$. – 2011-10-06
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5Sometimes, you'll see $\Re z$ and $\Im z$ used instead of $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$. – 2011-10-06
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4As 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. – 2011-10-06
2 Answers
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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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2If 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. – 2011-10-06
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1@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 $z$. – 2011-10-06
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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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0Do ALL real numbers get mapped on to the unit circle in the complex plane (or the Argand plane)? – 2011-10-06
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0@Tyler: No, only those numbers with an absolute value of **1** can lie in the **unit** circle. – 2011-10-06
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0@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. – 2011-10-06