-1

Here are two limits with different answers provided by Wolfram|Alpha:

Limit 1:

$$\lim_{z\to \frac{2}{5}-\frac{i}{5}} \frac{2\left(z-\frac{2-i}{5}\right)}{(1-2i)z^2+(6i)z+(-2i-1)}=-\frac{i}{2}$$

Limit 2:

$$\lim_{z\to \frac{2}{5}-\frac{i}{5}} \frac{2\left(z-\frac{2-i}{5}\right)}{\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right)}=-1-\frac{i}{2}$$

Aren't these the same limits because $(1-2i)z^2+(6i)z+(-2i-1)=\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right)$?

Here is what Wolfram|Alpha suggests.

projectilemotion
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Alex.F
  • 359

1 Answers1

1

The two denominators are not equal : notice that they don't have the same coefficient for $z^2$. Instead, you should have (according to the $3$rd link) the equality $$(1-2i)z^2+(6i)z+(-2i-1)=(1-2i)\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right),$$ so the two limits only differ from a factor $1-2i$. Then the results become coherent, as indeed, $$-\frac{i}{2}\left(1-2i\right)=-1-\frac{i}{2}.$$

Arnaud D.
  • 20,884
  • Thanks for the reply! I'm a little confused as to why that equality you've written is true. If the roots of $(1-2i)z^2+(6i)z+(-2i-1)$ are $(2-i)/5$ and $(2-i)$ then can't we just factorise it as I've written in the first post? Where does the $(1-2i)$ come from? – Alex.F Mar 19 '17 at 22:01
  • Your factorisation is wrong, because you have forgotten to take into account the leading coefficient. This leading coefficient is the $(1-2i)$. – Arnaud D. Mar 19 '17 at 22:07
  • Sorry for being a bit slow but it's not quite clicking. Is it not true that if you find the roots of $f(z)$ to be $z=a$ and $z=b$, then you may write $f(z)=(z-a)(z-b)$? In this case with $f(z):=(1-2i)z^2+(6i)z+(-2i-1)$, $a=(2-i)/5$ and $b=(2-i)$. I see that the $(1-2i)$ term is missing but don't understand why? – Alex.F Mar 19 '17 at 22:16
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    @Alex.F: the function $f(z)$ and $2\cdot f(z)$ have the same roots. Are you saying that they necessarily equal? – Willie Wong Mar 20 '17 at 01:52
  • @WillieWong Ah I think I get it now. So if $f(z)$ has roots $a$ and $b$ then $f(z)=c(z-a)(z-b)$ for some multiplicative constant $c$. Usually $c=1$ but not always, which is what I misunderstood. Right? – Alex.F Mar 20 '17 at 14:09
  • @Alex.F I wouldn't really say "usually", but that's right. – Arnaud D. Mar 20 '17 at 18:41