I tried to solve this question but the final solution which I obtain is not the as same as in the text book
Find domains of convergence of the series
$$\sum_{n=1}^ \infty \frac{1.3.5...(2n-1)}{n!}(\frac{1-z}{z})^n$$
The book answer is $|z-\frac{4}{5}|\lt \frac{2}{3}$
But my answer is $$|\frac{1-z}{z}|\lt \frac{1}{2}$$
Is this true ??
How can I obtain the solution as in the book .
On
@DonAntonio Your solution is correct :
$$\left|\frac{1-z}z\right|<\frac12\iff|2-2z|<|z|$$
Now put $\;z=x+iy\;,\;\;x,y\in\Bbb R\implies 2-2z=(2-2x)-2yi\;$ , so::
$$|2-2z|^2<|z|^2\iff(2-2x)^2+4y^2<x^2+y^2\iff$$
$$4-8x+4x^2+4y^2<x^2+y^2\iff3x^2-8x+3y^2+4<0\iff$$
$$3\left(x-\frac43\right)^2-\frac{16}3+3y^2+4<0\iff 3\left(x-\frac43\right)^2+3y^2<\frac43\iff$$
$$\left(x-\frac43\right)^2+y^2<\left(\frac23\right)^2\ldots$$
Thus
$$|z-\frac43|\lt \frac23 $$
Your solution is correct :
$$\left|\frac{1-z}z\right|<\frac12\iff|2-2z|<|z|$$
Now put $\;z=x+iy\;,\;\;x,y\in\Bbb R\implies 2-2z=(2-2x)-2yi\;$ , so::
$$|2-2z|^2<|z|^2\iff(2-2x)^2+4y^2<x^2+y^2\iff$$
$$4-8x+4x^2+4y^2<x^2+y^2\iff3x^2-8x+3y^2+4<0\iff$$
$$3\left(x-\frac43\right)^2-\frac{16}3+3y^2+4<0\iff 3\left(x-\frac43\right)^2+3y^2<\frac43\iff$$
$$\left(x-\frac43\right)^2+y^2<\left(\frac23\right)^2\iff\left|z-\frac43\right|<\frac23$$