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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!2012-11-10

8 Answers 8

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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!2012-11-10
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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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    @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.2012-11-11
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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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    This should be the accepted answer. Straightforward and clear.2014-01-19
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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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In this case $z = 1$ is a hole!

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    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 then2012-11-10
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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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    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.2012-11-10
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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.