If we were given a few conditions on a complex $z$, say "The modulus of $z$ is $20$" and "the argument of $z$ is $\pi/3$", its clear that $z$ is unique. To find $z$ we might start by writing "Let $z=a+bi$" then using the information from statement $1$ we can write $a^2 + b^2 = 400$ and from statement $2$ we can get $tan^{-1}({\frac{b}{a}})$ $=$ $\pi/3$. Then we can solve for $a,b$ etc. In this context are $a,b$ constants or variables? I guess my confusion stems from something more basic, is $x$ constant or variable in $(x-3)^2 = 0$? Solving gives $x=3$ and we know that $3$ is the only value for which the statement is true.
2026-03-28 15:26:48.1774711608
What is the meaning behind "Let" in this instance?
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Context is key here. Using the more simple example you posed: if we look at $(x-3)^2$ (could look at is as $(x-3)^2 = r$, where $r\in\mathbb{R}$), $x$ is a variable. When you set that expression equal to something, $(x-3)^2 = 0$, that's when the variable stands in for a constant.
Going back to the first question, $tan^{-1} \frac{a}{b} = \text{arg}$ and $a^2 + b^2 = \text{mod}^2$. $a$ and $b$ are variables. But when you have specific values for the "other side" of the equation, that's when your variables stand in for specific values (constant numbers in this cases).
Per the definition in wikipedia now (hopefully this provides another useful way of looking at the above answer):
Addendum: saw a really good comment and wanted to add. The way that variables and constants come about in this context is because people usually want to add some implicit context to the problem. For example, if we say that in $x+2$, $x$ is a constant, then we are assuming that $x+2$ always has the same value under any conditions. We are just not writing the explicit value of $x$ because it may be some complex number, i.e., 1.2494374 etc.
But when you say that $x$ is a variable, then we understand that $x$ has a wide range of possible values that can be affected by some other conditions, such as by us knowing that on friddays, $x+2 = 8$ but on mondays $x+2 = 5$.