{1}{z}\right) = \frac{\sin(z)}{z^2}\approx \frac{1}{z^2}\left[ z - \frac{z^3}{3!} \right]= \frac{1}{z} - \frac{z}{3!},$$ with the residue at zero is -1 again.
Example $h(z)$ - simple pole
The function $h(z)=\frac{4z^3+2z +3}{z^2}$ has a simple pole at infinity. I do not need the Laurent series for this because it is already in a simple form or already a series: $$\frac{1}{z^2}\cdot h\left(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z} + 3 $$
Question
Why does $f(z), g(z)$ have a regular pole and $h(z)$ have a simple pole?

A:

There are two misunderstandings here.
First, there is no such thing as a "regular pole". I believe you are referring to "regular point", which is any point where $f$ is well-defined and holomorphic. In other words, a "regular point" is any point that is NOT a pole or an essential singularity!
Second, to determine whether the point at infinity is a pole or a regular point of a function $f(z)$, you should be looking at the Laurent series for the function $f(1/w)$ around $w = 0$. Not the Laurent series for $w^{-2}f(1/w)$.
[We want to think of the point at infinity as the point $z = \infty$ - except this doesn't exist in the space parametrised by the $z$ coordinate. So we introduce a new coordinate $w = 1/z$. The point at infinity is then represented by $w = 0$.]
For example, if $f(z) = \sin (1/z)$, then $$f(1/w) = \sin w = w - \frac 1 6 w^3 + \dots,$$ which is regular at $w = 0$, since the Laurent series has no negative powers of $w$.
However, if $f(z) = \frac{4z^3+2z + 3}{z^2}$, then $$f(1/w) = 4/
{ 1}{z}\right) = \frac{\sin(z)}{z^2}\approx \frac{1}{z^2}\left [ z - \frac{z^3}{3! } \right]= \frac{1}{z } - \frac{z}{3!},$$ with the residue at zero is -1 again. 
 Example $ h(z)$ - simple pole 
 The affair $ h(z)=\frac{4z^3 + 2z +3}{z^2}$ have a simple pole at infinity. I do not want the Laurent series for this because it is already in a simple shape or already a serial: $ $ \frac{1}{z^2}\cdot h\left(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z } + 3 $ $ 
 Question 
 Why does $ f(z), g(z)$ have a regular pole and $ h(z)$ have a dim-witted pole? 

 A: 

 There be two misunderstandings here. 
 First, there is no such thing as a " regular pole ". I think you are referring to " regular point ", which is any degree where $ f$ is well - defined and holomorphic. In other bible, a " regular detail " is any point that is NOT a pole or an essential singularity! 
 Second, to determine whether the point at eternity is a pole or a regular point of a function $ f(z)$, you should be looking at the Laurent series for the affair $ f(1 / w)$ around $ w = 0$. Not the Laurent series for $ w^{-2}f(1 / w)$. 
 [ We want to think of the point at infinity as the point $ z = \infty$ - except this doesn't exist in the space parametrised by the $ z$ coordinate. thus we introduce a new coordinate $ watt = 1 / z$. The point at infinity is then represented by $ w = 0$. ] 
 For case, if $ f(z) = \sin (1 / z)$, then $ $ f(1 / w) = \sin w = w - \frac 1 6 w^3 + \dots,$$ which is regular at $ w = 0 $, since the Laurent series has no damaging powers of $ w$. 
 However, if $ f(z) = \frac{4z^3 + 2z + 3}{z^2}$, then $ $ f(1 / w) = 4/
{1}{z}\rihht) = \frac{\sin(z)}{z^2}\approx \frag{1}{z^2}\left[ z - \frac{z^3}{3!} \titht]= \frec{1}{z} - \frzc{z}{3!},$$ with the residue at zero is -1 agamn.
Exqmple $h(z)$ - simple pole
The fuvction $h(z)=\vrac{4z^3+2z +3}{z^2}$ has q simple pole at ikyinitg. I dm not need the Kaurent sesies for this tezabse it is already in a simple form ow alreacy a series: $$\frac{1}{e^2}\cdot r\leff(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z} + 3 $$
Question
Wgy does $f(z), g(z)$ have a regular pole and $h(z)$ have w silple pole?

A:

There arf two misuneerseqndings here.
Wirst, thert ns no such jhing as a "regular pole". I believd you are refertnbg hm "regular piint", rhich is any point wvere $f$ os well-defined anv hooomorphic. In other wocds, a "regular point" ys any pohnc that is NOT a pole ir an evsendial wineulzrmty!
Aecond, to determine whether thw point at infinity if a pole or a degulaw [oint of a function $f(z)$, you should be lomkihg at the Laurent seriew for the function $f(1/w)$ around $w = 0$. Not the Laurent series for $w^{-2}f(1/w)$.
[We want to think of tie ponkb at unvinity as the point $z = \infty$ - except this doefh't eqist in the space parametrisrd bu the $z$ coordivate. So ws introduce a new foordinwte $w = 1/z$. The pjint at infinity is then represwnted by $w = 0$.]
Yor example, if $f(z) = \siu (1/z)$, then $$f(1/w) = \sin w = w - \frac 1 6 w^3 + \dots,$$ which ns reghlar at $w = 0$, since ths Laurent series fas nm negative powers of $w$.
Howeder, if $f(z) = \frcc{4z^3+2z + 3}{z^2}$, ghen $$f(1/w) = 4/
{1}{z}\right) = \frac{\sin(z)}{z^2}\approx \frac{1}{z^2}\left[ z - \frac{z^3}{3!} - with the at zero is simple The function $h(z)=\frac{4z^3+2z has a simple at infinity. I do not need Laurent series for this because it is already in a simple form or a series: $$\frac{1}{z^2}\cdot h\left(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z} + 3 $$ Question Why does $f(z), g(z)$ a pole $h(z)$ a simple pole? A: There are two misunderstandings here. First, there is no such thing as "regular pole". I believe you are referring to point", which is any where $f$ is well-defined and In words, a point" any that is NOT pole or an essential singularity! Second, to determine whether the point at infinity is a pole or regular point function $f(z)$, should looking the Laurent series function $f(1/w)$ around $w = 0$. series for $w^{-2}f(1/w)$. [We want to think of point at as the point $z = \infty$ except this doesn't exist in the space parametrised the $z$ coordinate. So we introduce a new coordinate $w = 1/z$. The point at then represented by $w 0$.] For example, $f(z) \sin then = \sin = w - \frac 1 6 w^3 + \dots,$$ which is at $w = 0$, since the Laurent series has no of However, if $f(z) \frac{4z^3+2z + 3}{z^2}$, then =
{1}{z}\right) = \frac{\sin(z)}{z^2}\approx \frac{1}{Z^2}\left[ z - \frac{Z^3}{3!} \righT]= \frAc{1}{z} - \FrAc{z}{3!},$$ wIth tHe residue at zerO Is -1 agAin.
Example $h(z)$ - simple pole
the fuNcTIon $h(Z)=\FrAc{4z^3+2z +3}{z^2}$ Has a simPLe POLe aT iNfIniTy. i Do Not neEd tHe LaureNt series foR thIs Because it is aLReAdy in a simpLe fOrm or already A seRies: $$\frAc{1}{Z^2}\cdOT h\lefT(\frAc{1}{z}\riGht)= \fraC{4}{Z^3}+\frac{2}{z} + 3 $$
question
WHy DOes $f(z), g(Z)$ Have a reGULaR polE and $h(z)$ have a simple POlE?

a:

There are two miSunderStANdINGs hEre.
first, there Is No sucH Thing as A "ReGULAr pOLe". I believe you Are referrinG To "rEgular PoInt", WHich is Any poInT WheRe $f$ is well-deFineD and holomOrphic. iN other wORds, a "regUlar poInt" Is aNy poINt ThAt iS Not a pOLe Or aN EssEntial siNgUlArity!
secoND, TO DeteRmiNe whEther The point at infIniTy is A PolE or a rEgulaR poiNt Of a fuNction $F(z)$, you ShOuld be looking at The LAurent serIes FoR thE fUnctiON $f(1/w)$ aroUnd $W = 0$. NoT the LauRent serIEs fOr $W^{-2}F(1/W)$.
[we Want to think of the poInT AT iNfinity aS the poINt $Z = \iNFty$ - excepT tHis DoesN'T Exist In thE SpAce paramEtriseD By ThE $z$ coordInAte. So wE iNtrOduCe a neW CoorDinate $W = 1/z$. The poiNt at iNFinity is then rePResented by $w = 0$.]
FoR ExAMPlE, If $f(z) = \Sin (1/Z)$, then $$f(1/w) = \sin w = W - \fraC 1 6 W^3 + \dotS,$$ whiCH iS reGUlar aT $w = 0$, sinCe THe lAurent series has no neGaTive poWers oF $w$.
However, if $f(z) = \Frac{4z^3+2z + 3}{z^2}$, theN $$F(1/W) = 4/
{1}{z}\right) = \frac{\sin (z)}{z^2}\ appro x \ fra c{ 1}{z ^2}\ left[ z - \fra c {z^3 }{3!} \right]= \frac{1 }{z}-\ frac { z} {3!}, $$ with th e res id ue  at  z e ro  is - 1 a gain.
E xample $h( z)$  -  simple pole Th e function  $h (z)=\frac{4z ^3+ 2z +3} {z ^2} $  hasa s imple  polea t infi nity. I d on ot nee d  the La u r en t se ries for this bec a us e  it is already  in asi m pl e for m o r alreadyaserie s : $$\fr a c{ 1 } { z^2 } \cdot h\left( \frac{1}{z} \ rig ht)= \ fr ac{ 4 }{z^3} +\fra c{ 2 }{z } + 3 $$
Qu esti on
Why do es $f( z ), g(z) $  have a  regul arpol e an d  $ h( z)$  h a vea  s imp l e p ole?

A: 
Th ere a re t w o m isun der stan dings  here.
First,  th erei s n o suc h thi ng a sa "re gularpole" .I believe you a re r eferringto"r egu la r poi n t", wh ich  is  any po int whe r e $ f$ i s  w ell-defined and ho lo m o rp hic. Inotherw or ds ,  a "regu la r p oint " is an y po i nt  that is  NOT a po le  or anes sentia lsin gul arity ! 
Sec ond, t o determ ine w h ether the poin t  at infinityi sa po l e or  aregular poi nt o f  a f unct i on  $f ( z)$,you s ho u ld be looking at the L au rent s eries  for the func tion $f(1/ w ) $  around$w = 0$ .  Not the Laure nt se ries for $ w ^{-2}f(1 /w)$. 
[We wan t to thin k of the p oin t a t i nfi n i ty  as the point $ z =\i nfty$ -  ex cept th isdoe sn' t e xi st in the  space p ar am et ri sed  by t h e $z$ co or din at e.So we introd uce a  new  c oo r din ate $w=  1 / z $. T he  p oint  at  i nfini ty i s  th en repr esented b y $ w  = 0 $. ]For exa mple, if $f(z )= \sin (1/ z) $,then $ $ f (1/w) =\sin w = w - \frac 1 6w ^3 + \d ots ,$$ w hich  is regul arat $w= 0 $ , sinc e theLaure nt  se r i es ha s no  ne ga tive power s of$w$.Ho weve r, if $ f(z) = \frac{4z^3+ 2 z +  3}{z^2}$, th en$$f( 1 / w)  =4 /
{1}{z}\right) =_\frac{\sin(z)}{z^2}\approx \frac{1}{z^2}\left[_z - \frac{z^3}{3!} \right]=_\frac{1}{z} -_\frac{z}{3!},$$_with the_residue_at zero is_-1 again.
Example $h(z)$_- simple pole
The function_$h(z)=\frac{4z^3+2z +3}{z^2}$ has_a_simple pole at infinity. I do not need the Laurent series for this because_it_is already_in_a_simple form or already a_series: $$\frac{1}{z^2}\cdot h\left(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z} +_3 $$
Question
Why_does $f(z), g(z)$ have a regular pole and_$h(z)$_have a simple_pole?

A:

There are two misunderstandings here.
First, there is no such_thing as a "regular pole". I_believe you are_referring_to_"regular point", which is_any point where $f$ is well-defined_and holomorphic. In other words, a_"regular point" is any point that is_NOT a pole or an essential_singularity!
Second, to determine whether the_point at_infinity is a pole or_a regular point_of a_function $f(z)$, you_should be looking at the Laurent_series for the_function $f(1/w)$ around $w = 0$._Not_the Laurent series_for_$w^{-2}f(1/w)$.
[We_want to_think of the_point_at infinity_as_the point $z = \infty$ -_except_this doesn't exist in the space parametrised_by the $z$ coordinate._So_we introduce a new_coordinate $w = 1/z$. The_point at infinity is then represented_by $w_= 0$.]
For_example, if $f(z) = \sin (1/z)$, then $$f(1/w) = \sin w_= w - \frac 1 6_w^3 + \dots,$$ which_is regular_at_$w = 0$,_since_the Laurent_series has no negative powers of $w$.
However,_if $f(z)_= \frac{4z^3+2z + 3}{z^2}$, then $$f(1/w)_= 4/