Question #50231
Let f(z) be an analytic function in the annulus 0 <|z| < R for some positive real number R,Whose laurent series (in this annulus) is given by
f(z) = n from -∞ to ∞ ∑ { (-1)^n / (n^2)! ] } . Z ^ { 5n - n^2 -1}
A)) What Kind of Singularity is z=0 for f(z) ?
B)) Compute integral on Curve for [ z ^ 24 . f(z) dz] ,where C is a counterclockwise simple path lying in the annulus enclosing z=0
C)) Calculate Res (f) in z=0
D)) Evaluate Integral on Curve for [ sin Z .f(z) dz] , where C : |z| = (R/2) oriented counterclockwise
Note : please not by limit
f(z) = n from -∞ to ∞ ∑ { (-1)^n / (n^2)! ] } . Z ^ { 5n - n^2 -1}
A)) What Kind of Singularity is z=0 for f(z) ?
B)) Compute integral on Curve for [ z ^ 24 . f(z) dz] ,where C is a counterclockwise simple path lying in the annulus enclosing z=0
C)) Calculate Res (f) in z=0
D)) Evaluate Integral on Curve for [ sin Z .f(z) dz] , where C : |z| = (R/2) oriented counterclockwise
Note : please not by limit
Expert's answer
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Comments
Dear vallle. Thank you for adding information.
Dear vallle. Thank you for correcting us, explanation on Res is placed at our site. Function z^24*f(z) has singularity 0. We cannot use only z^24, because it is a multiplier, but we consider the whole function z^24*f(z).
Dear vallle. We added some explanations to the solution, please read them at our site.
Dear vallle. We added some explanations, read the solution at the site.
please in (B) n^2 -5n -24 = 0 not n^2 -5n +24 = 0 so the solution not zero thanks
D: please i need clear expansion by substitute n values please Also How use K,n ,and equation (3) please i don't understand this
A: I need the expansion by substitute n values please ,, How Z^24 . f(z) = n^2 - 5n -24+1=1 B -Please I need the general rule of this or support this by another clear method What D How D=25-4.4 Then how Res 0 ,you mean No Singularity inside Curve for f (z) ,please explain more ? So what about Z^24 i think have singular at z=0 Right? so why we cannot only use Res 0 it in integral C:also How use 5n - n^2 = 0 please i need clear expansion of (1) to explain the coefficient exactly by substitu n
b branch : i think it's wrong How the Res is zero what about Z^24 i think have singular at z=0 Right? so why we cannot only use Res 0 it in integral