Hi please could you assist me: questions posted below:

Assuming the function f is holomorphic in the disk $D(0,1) = \{ z \in \mathbb{C}:|z| < 1\}$, prove that $g(z) = \overline {f(\overline z )}$ is also holomorphic in D(0,1) and find its derivative?????

Find the radii of convergence of the following series stating which result is being used.
(a) $\sum\limits_{k = 0}^\infty {k^{113} 2^{ - k} (z - 1)^k }$

(b) $\sum\limits_{n = 2}^\infty {n!(z - e)^{3n} }$

(c) $\sum\limits_{k = 0}^\infty {\frac{{z^k }}{{(k!)^2 }}}$

(A)
Using Ratio Test:

$\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
{{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - (k + 1)} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| =$

$
\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - k - 1} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - 1} (z - 1)}}{{k^{113} }}} \right|$

= $\frac{{(\infty + 1)^{113} 2^{ - 1} (z - 1)}}{{\infty ^{113} }} = \frac{{(z - 1)}}
{2} < 1$

So is rad. of convergence (z-1)/2 and converging since it is less than 1??

(B)

Using Ratio Test:

$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
{{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)!(z - e)^{3(n + 1)} }}{{n!(z - e)^{3n} }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {(n + 1)(z - e)^3 } \right|$

$= (\infty + 1)(z - e)^3 = \infty$

Hence, is ROC infinity and diverging since it is greater than 1?

(C)
Using Ratio Test:

$\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
{{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{((k + 1)!)^2 }} \times \frac{{(k!)^2 }}
{{z^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{z^k }} \times \frac{{(k!)^2 }}
{{((k + 1)!)^2 }}} \right|$

$= \mathop {\lim }\limits_{k \to \infty } \left| {z \times \frac{{(k!)^2 }}{{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\sqrt z \frac{{k!}}{{k + 1!}}} \right|$

$= \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{\sqrt z }}
{{k + 1}}} \right| = \frac{{\sqrt z }}{{\infty + 1}} = 0$

So is ROC 0 and converging since it is less than 1?

2. Originally Posted by rinatoc

Assuming the function f is holomorphic in the disk $D(0,1) = \{ z \in \mathbb{C}:|z| < 1\}$, prove that $g(z) = \overline {f(\overline z )}$ is also holomorphic in D(0,1) and find its derivative?????

using complex conjugate properties and this fact that $z \rightarrow a$ iff $\overline{z} \rightarrow \overline{a},$ we get:

$\lim_{z\to a} \frac{g(z)-g(a)}{z-a}=\lim_{z\to a} \frac{\overline{f(\overline{z})} - \overline{f(\overline{a})}}{z-a}=\lim_{\overline{z}\to \overline{a}} \overline{\left(\frac{f(\overline{z})-f(\overline{a})}{\overline{z}-\overline{a}} \right)}$

$=\overline{\left(\lim_{\overline{z}\to \overline{a}} \frac{f(\overline{z})-f(\overline{a})}{\overline{z}-\overline{a}} \right)} \ \ \ \ \ \text{since the map} \ \ z \rightarrow \overline{z} \ \ \text{is continuous}$

$=\overline{f'(\overline{a})}.$ thus $g$ is holomorphic too and $g'(z)=\overline{f'(\overline{z})}. \ \ \ \square$

Find the radii of convergence of the following series stating which result is being used.
(a) $\sum\limits_{k = 0}^\infty {k^{113} 2^{ - k} (z - 1)^k }$

(b) $\sum\limits_{n = 2}^\infty {n!(z - e)^{3n} }$

(c) $\sum\limits_{k = 0}^\infty {\frac{{z^k }}{{(k!)^2 }}}$

(A)
Using Ratio Test:

$\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
{{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - (k + 1)} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| =$

$
\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - k - 1} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - 1} (z - 1)}}{{k^{113} }}} \right|$

= $\frac{{(\infty + 1)^{113} 2^{ - 1} (z - 1)}}{{\infty ^{113} }} = \frac{{(z - 1)}}
{2} < 1$

So is rad. of convergence (z-1)/2 and converging since it is less than 1??[/tex]

you should say that $\lim_{k\to\infty} \frac{(k+1)^{113}}{k^{113}} = 1,$ rather than putting $k=\infty$!!!

so you got $|z-1| < 2,$ which means that the radius of convergence is 2.

(B)

Using Ratio Test:

$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
{{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)!(z - e)^{3(n + 1)} }}{{n!(z - e)^{3n} }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {(n + 1)(z - e)^3 } \right|$

$= (\infty + 1)(z - e)^3 = \infty$

Hence, is ROC infinity and diverging since it is greater than 1?

so the series is divergent for all values of z except for $z=e.$ thus the radius of convergence is 0.

(C)
Using Ratio Test:

$\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
{{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{((k + 1)!)^2 }} \times \frac{{(k!)^2 }}
{{z^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{z^k }} \times \frac{{(k!)^2 }}
{{((k + 1)!)^2 }}} \right|$

$= \mathop {\lim }\limits_{k \to \infty } \left| {z \times \frac{{(k!)^2 }}{{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\sqrt z \frac{{k!}}{{k + 1!}}} \right|$

$= \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{\sqrt z }}
{{k + 1}}} \right| = \frac{{\sqrt z }}{{\infty + 1}} = 0$

So is ROC 0 and converging since it is less than 1?

where did $\sqrt{z}$ come from??! since $(k+1)!=(k+1)k!,$ you'll get:

$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\infty.$

3. Originally Posted by NonCommAlg
so the series is divergent for all values of z except for $z=e.$ thus the radius of convergence is 0.
i understand your method here, thanks for that. but why is it 0.
is there a general rule saying that the ROC is 0 if the function is divergent everywhere except for one point?

Originally Posted by NonCommAlg
where did $\sqrt{z}$ come from??! since $(k+1)!=(k+1)k!,$ you'll get:

$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\infty.$
lol i was trying to square root the z so that it would be easier to work with the k's but your method is more appropriate.
as before though, why is the ROC infinity. is there a general rule saying that if you get the value as 0 then the ROC is infinity?
how did you conclude these answers?

thanx very much for the help btw, very much appreciated!!!