- #1
nhrock3
- 415
- 0
find the laurent series of [tex]f(x)=\frac{-2}{z-1}[/tex]+[tex]\frac{3}{z+2}[/tex]
for
1<|z|<2
i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
[tex]f(x)=\frac{1}{1-z}[/tex]
then
the radius is 1 and
because 1-1=0
so
it is analitical on
|z|<1
so if i apply the same logic
[tex]f(x)=\frac{-2}{z-1}[/tex]
1 still makes denominator 0
and
it is analitical on
|z|<1
but the correct answer is
it is analitical on
|z|>1
for
[tex]f(x)=\frac{3}{z+2}[/tex]
-2 makes denominator 0
so |z|<-2 (but its illogical because |z| is a positive numbe)
where is my mistake?
for
1<|z|<2
i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
[tex]f(x)=\frac{1}{1-z}[/tex]
then
the radius is 1 and
because 1-1=0
so
it is analitical on
|z|<1
so if i apply the same logic
[tex]f(x)=\frac{-2}{z-1}[/tex]
1 still makes denominator 0
and
it is analitical on
|z|<1
but the correct answer is
it is analitical on
|z|>1
for
[tex]f(x)=\frac{3}{z+2}[/tex]
-2 makes denominator 0
so |z|<-2 (but its illogical because |z| is a positive numbe)
where is my mistake?