Question

residue

Answer 1

\[\frac{ e^{iz} }{ (z^2+4)^2 }\]

Answer 2

at z=2i. The order of pole is 4

Answer 3

\[\frac{ 1 }{ 3! }\frac{ d^3 }{ dz^3 }(e^{iz})|z=2i\]

Answer 4

\[\frac{ -ie^{-2} }{ 6 }\]

Answer 5

@Michele_Laino

Answer 6

we can rewrite such function as below:
\[\Large \begin{gathered}
f\left( z \right) = \frac{{{e^{iz}}}}{{{{\left( {{z^2} + 4} \right)}^2}}} = \frac{{{e^{iz}}}}{{{{\left( {z - 2i} \right)}^2}{{\left( {z + 2i} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
so, I think that \(z=2i\) is a pole of order \(2\)

Answer 7

Would it be wrong if I didn't realize that it is order of 2 and not expand?

Answer 8

yes! of course

Answer 9

How to realize when to expand and when not to?

Answer 10

usually, when I have to do such computation, I always write the factorization the denominator

Answer 11

like when I've sin and cos I'll use taylor series and cancel out the denominator to identify the order. If in a case it would cancel will that be the power of the function they specified in the problem. In other words, I need to simplify

Answer 12

if we have a sin or cos function, I think it is better to use these substitutions:
\[\Large \cos z = \frac{{{e^{iz}} + {e^{ - iz}}}}{2},\quad \sin z = \frac{{{e^{iz}} - {e^{ - iz}}}}{{2i}}\]

Answer 13

Got you thanks

Answer 14

:)