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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.

Nav Bhatthal
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    I have never understood the need for some teachers and textbook authors to make a strict distinction between "variables" and "constants" in this manner. – Arthur Sep 02 '23 at 12:01
  • @Arthur does that mean you would say $x$ is a variable in $x+1=2$? Just because its a letter and not a number such as $2$ or $100000\pi$? – Nav Bhatthal Sep 02 '23 at 12:02
  • Related, possibly helpful: https://math.stackexchange.com/questions/2738360/what-exactly-is-an-equation/2738382#2738382 – Ethan Bolker Sep 02 '23 at 12:16
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    Please see Q859288 and the linked pages there. "Variables" and "constants" (and "parameters", etc.) are more about what we do with the term than what the term is. In $f(x)=ax^2+3$, we treat each of $a, x, 3$ differently in terms of how we change it and how we define other terms with it. We would tend to call $x$ a "variable" and $3$ a "constant". And we while we would tend to call $a$ a variable, if we set $a=5$ and never change it, shouldn't it be a constant? The words aren't strict and all of the above are "terms". – Jam Sep 02 '23 at 12:27
  • How does this relate to my example? What do you think the most commonly used term for $a,b$ are (in my context)? Thanks for the reading material – Nav Bhatthal Sep 02 '23 at 12:29
  • If $(x-3)^2 = 0$ is a given, then $x$ is a constant (it's fine to call it an unknown even if some of us do know that it equals $3$). Otherwise, we can call $x$ a variable that makes $(x-3)^2 = 0$ true at particular values. Whatever we choose to call $x$ really depends on the context. – ryang Sep 02 '23 at 17:49

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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):

a variable is a symbol that represents a quantity in a mathematical expression

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 $x+2 = 5$.

ning
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  • Once we have the condition's for $a,b$, that is the length of $z$ and its arg, there is a "mental shift" in which we now think of $a,b$ as being constant? Thats correct right? I dont agree with your thing about $r$ though, if we sub in a value for $r$, $x$ becomes constant, I think $f(x) = (x-3)^2$ is the variable interpretation. – Nav Bhatthal Sep 02 '23 at 12:06
  • I like your use of notation here more for $f(x) = (x-3)^2$ instead of $(x-3)^2$ for some r. But I'd say that your mental shift is shifting towards the right direction. I also just edited my answer a bit more because I saw arthur's comment I and I thought there was a hidden gem posed in that comment. – ning Sep 02 '23 at 12:14
  • When we go from $f(x) = (x-3)^2$ to $(x-3)^2 = 0$ $(1)$ we do not bother to change the letter, i.e we could say $\alpha$ is the value of $x$ such that $(1)$ is true, then work out that $\alpha = 3$, is that because its usually clear to the reader what the author means by "setting equal to $0$"? (and that means looking at a particular value(s) of $x$ for which $(1)$ is true) – Nav Bhatthal Sep 02 '23 at 12:17
  • I think you go it. I don't know if this other way of looking at it helps but in the example you provide, with the expression $(x-3)^2$ as a variable, we can find what $x$ would have to be for any value of $f(x)$ that we are interested. Otherwise, to explore $f(x)$ we would need a graph or a table of values. – ning Sep 02 '23 at 12:36