Question

(6) Show that all singularities of f(2) and evaluate sin() are simple poles. Find the residues 22:-1 ffo where is the circle 2| = 3 in ced.

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Answer 1

Residue of simple pole at h=9 G-a) frz) So bere Residne loro ZKI i (Z-kii) ana SINKTI Note calculation) Residne Tbeorens If fca) is analytie inside and on an closed contout except for a fimte number of isolated singularities inside or, lben fcanda (sura Insider Y al um of all residue are Here 8: 121=3, ie crocle conleord at on origin and radius 3 singulariter KTI ( O, Ti, ani....) only ő is the ő is the sogularity which lies inside. the cradle 171=3. s fcanda= ani x Residne at zero

lins (2-0) Sloz sdne at o % from) Zx Coszt Sanzxl 7-) .22 Ch Hopitals onle) Oxos o t Sino low feh = s to alix sum of Residne - 121=3 =an1X I 2 - Ti

Given function is fa- Sint ean_ I Singularities are given by eaz_1=0 ie., 2=1 2(x+iy) z=xrig) (since 22 — 2x+214 ) ie, ein) = 1 Now Bryal -1 =7 Cos cosay tisinay ie y= KI ( since even molteple of Jos y equals zero =xtiy -oti KI Kri the singularities. Note :- condition for ebeeking poles by lim (z-a) fc 7 is Since are is given Disite Za Now Irm KTR (Z-KII) Sinz lim (Z-Ktil Z KTT 21 x/lim Sinz スコKT1 e27-1

=lim Z-Kti x line Sio Z Zokni -- ZKTII % SIRKTI Sinz x lim Zokni Z-)KTIT 23. 1x Slokni finite singularities are poles and it is of order 15e poles are simple. one Reason: 7=a is a pole order K it lim スーつの of CZ-ajt fra) = finite (40 and a) (7- ki) 5107 Here lim Z- Ti is finite tn_1 so simple poles

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