1. radius of convergence question..(complex function)

i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
$f(x)=\frac{1}{1-z}$
then
because 1-1=0
so
it is analitical on
|z|<1

so if i apply the same logic
$f(x)=\frac{-2}{z-1}$
1 still makes denominator 0
and
it is analitical on
|z|<1
it is analitical on
|z|>1

for
$f(x)=\frac{3}{z+2}$
-2 makes denominator 0
so |z|<-2 (but its illogical because |z| is a positive numbe)

where is my mistake?

2. Originally Posted by transgalactic
i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
$f(x)=\frac{1}{1-z}$
then
because 1-1=0
so
it is analitical on
|z|<1

so if i apply the same logic
$f(x)=\frac{-2}{z-1}$
1 still makes denominator 0
and
it is analitical on
|z|<1
it is analitical on
|z|>1

for
$f(x)=\frac{3}{z+2}$
-2 makes denominator 0
so |z|<-2 (but its illogical because |z| is a positive numbe)

where is my mistake?
Your logic is a bit messy... Let's clean this up a bit.

Remember that for a geometric series

$S_n = a + ar + ar^2 + ar^3 + \dots = \sum_{n = 0}^{\infty}ar^n$, we can write this in closed form as

$S_n = \frac{a}{1 - r}$ when $|r| < 1$.

If you have $f(z) = \frac{1}{1 - z}$, then we can say $a = 1, r = z$.

So $f(z) = 1 + z + z^2 + z^3 + \dots = \sum_{n = 0}^{\infty}z^n$.

This is convergent for $|r| < 1$, so $|z| < 1$.

The radius of convergence is $1$.

Can you follow some similar logic for Questions 2 and 3?

3. when i apply your word there i get the same answer as before
could you do the others too?

4. No, you have a try.

The second is almost as easy as the first, simply apply a transformation so that it is of the form $\frac{a}{1 - r}$. At the moment the denominator is $r - 1$... What can you do to turn it into $1 - r$?

On second viewing, I get the same answer as you. The answer given is wrong, since

$f(z) = \frac{-2}{z - 1} = \frac{-2}{-(1 - z)} = \frac{2}{1 - z}$.

So $a = 2$ and $r = z$.

Therefore $f(z) = 2 + 2z + 2z^2 + 2z^3 + \dots = \sum_{n = 0}^{\infty}2z^n$, which is convergent for $|r| < 1$. So $|z| < 1$. The radius of convergence is $1$.

5. $f(x)=\frac{-2}{z-1}$
$f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}$
i cant getto tht form that is the best i got
??

6. $f(x)=\frac{-2}{z-1}$
$f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}$
i cant got to that form, that is the best i got
??

7. Originally Posted by transgalactic
$f(x)=\frac{-2}{z-1}$
$f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}$
i cant got to that form, that is the best i got
??
See the edit on my previous post for Question 2.

For Question 3 you have

$f(z) = \frac{3}{z + 2} = \frac{3}{1 + z + 1} = \frac{3}{1 - (-z - 1)}$.

What do you think $a$ and $r$ are?

8. but for question two the answer is|z|>1 not
|z|<1

For Question 3, I have already put the expression into $\frac{a}{1 - r}$ form for you. Surely you can read off $a$ and $r$, and be able to solve $|r| < 1$ for $z$...