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In the equation: $$\frac{z^2-1}{z-1}$$ $z$ can not be equal to $1$.
However $$\begin{align} \frac{z^2-1}{z-1}&=\frac{(z-1)(z+1)}{z-1}\\ &=(z+1) \end{align}$$ So then if $z$ is equal to $1$ we have $$\frac{z^2-1}{z-1}=2$$

Can someone explain that please?

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You have just solved a limit without even knowing it! –  Joel Nov 10 '12 at 23:27
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8 Answers

up vote 5 down vote accepted

The point at $z = 1$ is called a "removable singularity", and this process "removes" it. In general, a function having a limit at a point but not defined there can be extended to a function continuous at that point in only one way. That's what happens here. $\frac{z^2 - 1}{z - 1}$ is defined and continuous everywhere except $z = 1$, and has a limit at $z = 1$. $z + 1$ is defined and continuous everywhere, and equals $\frac{z^2 - 1}{z - 1}$ wherever both are defined. So then it must be that unique extension.

NB: It's worth noting that as other posters have pointed out, this equation is not technically correct since one expression is undefined at the point and the other isn't, but the procedure you used gives the unique extension just mentioned.

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Rather than making a statement like \begin{align*} \frac{z^2 - 1}{z-1} = z + 1 \end{align*} without saying what $z$ is, you should make a more careful statement like this:

If $z \in \mathbb{R}$ and $z \neq 1$, then $\frac{z^2 - 1}{z - 1} = z + 1$.

It wouldn't make sense to plug in $z = 1$, because that equation is not even true if $z = 1$.

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Algebraists will talk about things like $\mathbb C(z)$, the "field of rational functions over $\mathbb C$", and in that field it is indeed the case that $\frac{z^2-1}{z-1} = z+1$, but this doesn't say anything about being able to evaluate the function defined by $f(z) = \frac{z^2-1}{z-1}$ at $1$. What it does say is that $z^2-1 = (z-1)(z+1)$.

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It is structures like $\mathbb{C} (z)$ that best capture our intuitions about manipulating algebraic expressions: there is an obvious sense in which $\frac{z^2 - 1}{z-1} = z + 1$ and it should not have to involve taking limits! –  Zhen Lin Nov 10 '12 at 20:28
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By writing $z-1$ in the denominator you are already implicitly assumthat $z≠1$. Even if you derive the expression $z+1$ or anything else from the previous one, the assumption still holds that $z≠1$.

What you showed above holds if and only if $z≠1$. Conversely, if $z=1$ it does not hold. In the case that $z=1$, the expression $\frac{z^{2}-1}{z-1}$ is undefined, so it cannot be reduced to anything.

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In this type of case, for any value of z other than 1, the terms will cancel out and the function will behave like z+1. Notice that when z=1, the function is undefined. That means that that value can't be in the domain. We call it a hole.

To try and understand why you still cannot divide by 0 even in cases like this, try to graph out the function $f(x) = \frac{1}{x}$ and plug in values near 0 from the left and the right. Try values in this interval $0 \leq x \leq 1$ and notice what happens as you get closer to 0. After you check this by hand, go to wolfram or your calculator and look at what happens to verify. I hope this helps.

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But it can be in the domain if you just simplify the equation, no? –  Gineer Nov 10 '12 at 16:44
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Well sure, but you have to treat this type of problem as a "I know what you were" I suppose. If I were to allow myself to simply divide it away I would lose the possibility of holes. Also, looking at the undivided part written out it is clear division by 0 could happen for a particular value of z. These types of functions are interesting, therefore we consider them in the way we described where we choose not to forget original candidates for discontinuity. –  Ezea Nov 10 '12 at 21:13
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Let $f(z) = \frac{z^2-1}{z-1}$ and $g(z) = z+1$ then $f$ is analytic on $\mathbb C \setminus \{1\}$ and $g$ on $\mathbb C$.

A theorem of complex analysis says that if two functions coincide on an open set then they are equal everywhere that they are both defined.

This implies that there is a unique way to continue a function (such as $f$) onto a larger domain if it shares some overlap with another function (such as $g$).

This justifies continuing $f$ to a function on the whole of $\mathbb C$ by defining $f(1) = 2$.

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Can you please explain what "analytic" means in this context? –  Gineer Nov 10 '12 at 16:46
    
@Gineer, it's a particularly good type of function that is studied in complex analysis. The point is, if we are only thinking about these types of functions there is a unique way to define $f(1)$. –  sperners lemma Nov 10 '12 at 16:54
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@spernerslemma I don't think the OP is referring to complex numbers with the symbol $z$. –  Nameless Nov 10 '12 at 17:37
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@Nameless: Even so, the argument carries over by replacing '$\mathbb{C}$' by '$\mathbb{R}$', 'analytic' by 'continuous' and ignoring the second paragraph (which is far more general a result than is needed). I +1'd this answer because it emphasises that $z \mapsto (z^2-1)/(z-1)$ and $z \mapsto z+1$ are different functions (in intension, at least). –  Clive Newstead Nov 10 '12 at 21:33
    
@CliveNewstead Of course it carries over to the real numbers. My problem is this answer seems to be well above the mathematical maturity level of the OP, so for him it is useless. –  Nameless Nov 11 '12 at 12:54
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This step $\begin{align} \frac{(z-1)(z+1)}{(z-1)}&=z+1 \end{align} $ is only valid when $z \not=1$.

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sigh...........?!? I'm sure if I understood math, I could actually love it, but so far it eludes me. :-( –  Gineer Nov 10 '12 at 16:20
    
But I could start with f(x) = x that is defined everywhere and expand it by x/x to f(x) = x^2/x. This should not be defined at x = 0 suddenly? Isn't it so that a function is defined everyhwere when the numerator has a higher power than the denominator? Or the other way round: If I take the limit z -> 1 in the question, I think this would be allowed and leead to the result 2. –  Foo Bar Nov 10 '12 at 16:53
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@Foo Bar If somebody defines a function by saying $f(x) = \frac{x^2}{x}$, then they haven't defined their function carefully because they haven't specified the domain of the function. But assuming they have not specified their domain, the convention is to assume they meant all real numbers where this expression is defined, which would not include $0$. –  littleO Nov 10 '12 at 21:35
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@Gineer think of learning math like learning swimming. You cannot fight the water, you gotta accept it for what it is and work with it. Math has a super steep learning curve and you gotta be patient and work through a lot of problems to get a sense of what might be going on. Solving lots and lots of problems yourself is the only way to learn math. –  Amatya Nov 11 '12 at 1:19
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This should be the accepted answer. Straightforward and clear. –  Pacerier Jan 19 at 4:25
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In this case $z = 1$ is a hole!

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So are you saying the denominator is equal to the enumerator and therefor it's just 1? In that case, it's still not equal to 2 though? –  Gineer Nov 10 '12 at 16:15
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@Gineer: No, LJym89 is saying that the function is undefined at $z=1$, so its graph has a hole there. –  Brian M. Scott Nov 10 '12 at 16:19
    
O yea... whole $\ne$ hole. I think its getting late. Thats for mentioning it though, makes sense if I think of it graphically... Is that true though, because if you simplify the equation z can be equal to 1. The two equations are equal to each other then –  Gineer Nov 10 '12 at 16:28
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