Question

Develop f(z)=1/(z(z-3)) in a laurent series valid for the indicated domains.

0시리 <3 6) 3<시리

determine the nature of the singularities of the following functions.

Answer 1

had ho regular part, in 1 f(2) has Principal part. & t (2) has essential singulenty at 2=0 in 3 21 121 Thes, Egho can be written as Ifras mo zht2 which is lourent series Expansion, Aus 5 & 712170 Z (2-3) Z-(2-3) (2) f 그 - 1 3 2-3 3 Z (2-3) ZE f (2) = (7-17 ] ) + (OF [7 - (2-1) & J & = (2) [z- fth + 2 + 4 + 1) F - J J = (F) 2) (2) + 23 ģ 22 27 81 - (3 2 tr -- 81 243 32 I Regular part

27 13 2 Note: Any function has only regular art in laurent series expansion, then function f(2) has removable singularity at that point where f(2) attains indeterminate form, the given function Jcz 22 e- Here, at zoo 2 fcol which is indeterminate D & Laurent series expansion of f(2)= 2.2 e I 2 22: e 1 xt 22 (22) ² 2! -) + z z 2 2 + 4 2 + 8 2 3) 21 22 They has removable at z=0. Z which is only regular part of 2.S.E. singularity our function is not analytic at 20. Thus, 2.0 is singular point. Ay

Ou; 1Pg. 3 in a laurent series Develop f2) = 2 (2) 3 L 1Z) 3 LI 121 . f (2) 1-3 2.2 (1 - 2 / 2 2² (1+(2+3)+(+j+-) 2 + 27 5 + + 3 23 3t 4 (2) Z Laurent Series

-2 sec(42-42 f (2) = 22 Poler Any function fry=&a is of the justient form. fey will have pole at these points where h(2) 20 is analytic at those points. They fyz secuz-42 2² g (2) h(2) where, g(2) = secuz-uz hozz2 h (2) 22 2=0 Thes, 2=0 is a pole order 2 [ign=secuzur is analytic at 250 has pole of order 2 ( degree Hence, secuzi-uz 22 order Aug order 2 f(2)=z=0; f'(2) = 2230; 2 0; f2) = 2 80 L f (2) = 2 80 +2 means

f (2) has pole order ! is simple pole at z=0 Pg. 5