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11:00 AM
Jyrki Lahtonen
has unfrozen this room.
Martin Sleziak
11:16 AM
The room was unfrozen after the request in the Math Mods' Office:
in Math Mods' Office,
4 hours ago
, by
BAYMAX
Can anybody deal with the frozen room of complex analysis ?please help
1 hour later…
BAYMAX
12:43 PM
\[ \left\{
\begin{array}{ll}
0 & z = 0 \ \\ \frac{1}{\exp{(z^4)}} & z \neq 0\ \end{array}
\right. \]
How can i show that its not continuous at $z = 0$
Martin Sleziak
Let's have a look at slightly simpler $1/\exp(z)$, I do not think there is much of a difference.
If I look only at complex numbers of the form $z=iy$ then $$\frac1{\exp z}= \frac1{\cos y + i\sin y}.$$
$$\lim_{y\to0} \frac1{\cos y + i\sin y} = 1 \ne 0.$$
So this function is not continuous at $z=0$.
Does this solve the problem @BAYMAX?
BAYMAX
Yes,yes .. here also we can take $z^4= w$ and proceed..
yes
1 hour later…
Simply Beautiful Art
2:00 PM
:P Hello
Martin Sleziak
Is it me you're looking for?
Simply Beautiful Art
Nah, it's cool
Just saying hi
Martin Sleziak
It was meant as a reference to
Lionel Richie song
which contains the lyrics "Hello, is it me you're looking for?" But maybe I am too old, which is why the attempt to make a joke was lost.
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