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Determine the following limit or explain why the limit in question does not exist. $$ \lim_{z \to 1+i} \frac{z^2 - 2z + 2}{\lvert z \rvert^2 - 2} $$ I found this question online and was wondering what the answer should be for this one, thanks!

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up vote 1 down vote accepted

The limit does not exist.

The denominator is zero everywhere on the circle of radius $\sqrt{2}$ centered at the origin, so the ratio $$ \frac{z^2 - 2z + 2}{\lvert z \rvert^2 - 2} $$ blows up in every neighborhood of the point $z=1+i$.

Next let $z=(1+i)(1+t)$ where $t$ is real and small. For $t \neq 0$ we have $$ \frac{z^2 - 2z + 2}{\lvert z \rvert^2 - 2} = \frac{-1+i(1+t)}{2+t} \longrightarrow \frac{-1+i}{2} $$ as $t \to 0$.

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Here's one way: let $z=(a+1)+(b+1)i$ and then check that what you want is the limit, if it exists, of $${a^2-b^2-2b\over a^2+2a+b^2+2b}$$ as $a$ and $b$ go to zero (and you really should check this, as I did the algebra in a hurry). Then see what happens if you fix $a=0$ and let $b$ approach zero, and what happens if you fix $b=0$ and let $a$ approach zero.

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