Find the radii of convergence of
1) \(\sum_{n=1}^\infty \dfrac{(-3i)^n}{n^3}z^n\)
2) \(\sum_{n=1}^\infty (\dfrac{2in +1}{3n -2i})^n z^n \)
Please, help

- Loser66

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- Loser66

@oldrin.bataku

- freckles

I haven't done radii of convergence with complex numbers before
but I think it could be similar
do you have the answers?

- freckles

\[|z|=|z i | \text{ since if } z=a+bi \text{ then } |z|=\sqrt{a^2+b^2} \text{ and } iz=ia-b \\ \text{ so } |iz|=\sqrt{(-b)^2+(a)^2}=\sqrt{b^2+a^2}=\sqrt{a^2+b^2} \\ \text{ so for the first one I think I can say } \\ |\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}|=|\lim_{n \rightarrow \infty} \frac{(-3i)^{n+1} z^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3i)^nz^n}| \\ =|\lim_{n \rightarrow \infty} (-3i)z (\frac{n}{n+1})^3| \\ =|(-3i)z (\lim_{n \rightarrow \infty} \frac{n}{n+1})^3| \text{ and we want this less than 1}\]
you know where we can say |-3iz|=|3iz|=3|iz|=3|z| I believe

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## More answers

- freckles

I think if I were you I would try the root test on the second one

- freckles

that seem to work
the key is knowing |iz|=|z|

- Loser66

I am sorry, my computer is lagging.

- Loser66

I don't have the answers for them.

- Loser66

In class, we use either \(R = \dfrac{1}{limsup \sqrt[n]{|a_n|^n}}\) Or \(R =\lim|\dfrac{a_n}{a_{n+1}}|\)
We don't put z inside the lim

- Loser66

And we didn't have any example of the radius involve to multiple of i like this.
We have |z -i| <1, that is z's are inside the circle center i with radius 1

- Loser66

As you stated above , the lim =1, hence we just consider \(|\dfrac{-1}{3iz}| <1 \)

- freckles

for the first one I get |z|<1/3
which means the radius is 1/3

- Loser66

That throws me off also.
We have R =\( lim |\dfrac{a_n}{a_{n+1}}|\)

- freckles

http://www.wolframalpha.com/input/?i=sum%28%28-3i%29%5En%2Fn%5E3*z%5En%2Cn%3D1..infty%29
I assumed I did right because wolfram agrees
but there have been times wolfram has been wrong

- freckles

it seems the other way you have it also works

- Loser66

I show you the book.

##### 2 Attachments

- Loser66

theorem 1.3 and proposition 1.4

- Loser66

##### 1 Attachment

- Loser66

To me, if I use limit comparison test, the result should be the same. But they are not, right? Ok, let check
if use the proposition in the book \(lim|\dfrac{a_n}{a_{n+1}}|= lim\frac|{(-3i)^n}{n^3}*\dfrac{(n+1)^3}{(-3i)^{n+1}}|=|\dfrac{1}{(-3i)}lim\dfrac{n^3}{(n+1)^3}=\dfrac{1}{|-3i|}\)

- freckles

\[|\frac{-1}{3iz}|<1 \\ \frac{|-1|}{|3iz|}<1 \\ \frac{1}{3|z|}<1 \text{ since } |iz|=|z| \\ \text{ multiply both sides by } |z| \\ \frac{1}{3}<|z| \\ \text{ hmm... I don't get why they put } a_{n+1} \text{ on the bottom } \\ \text{ \because this should be } |z|<\frac{1}{3} \text{ which says the radi is } \frac{1}{3}\]

- Loser66

So the radius of convergence (if it works) is 1/3
while limit comparison test you post above is z > 1/3

- Loser66

They are|dw:1444010175317:dw| completely different , right?

- freckles

|z|>1/3 is definitely different from |z|<1/3
we should be getting |z|<1/3 which is what I obtained from doing |a_(n+1)/a_n|<1
but doing the book's way gives |z|>1/3
unless we are suppose to solve |a_n/a_(n+1)|>1 instead

- Loser66

hey, the book's solution give me it is < 1/3

- freckles

http://math.stackexchange.com/questions/378181/radius-of-convergence-of-power-series-of-complex-analysis
here they also put the n+1th on top

- Loser66

I know, but you tell me, should I follow my prof's and my book??? hehehe

- freckles

if you got the book's way gives |z|<1/3
then did you solve |a_n/a_(n+1)|>1

- Loser66

Nope, the concept is: if we calculate directly R by that method, that means on the radius R, the series converges (not including the boundary) |dw:1444010543836:dw|

- Loser66

At the point on the boundary (the circle) the series may or may not converge.

- Loser66

Because inside is convergent, outside is divergent, right on boundary : no information.

- freckles

There are the exact ways I would have done the problems:
we are doing n goes to infinity
so I will probably not write the word limit...
\[|\frac{a_{n+1}}{a_n}|<1 \\ |a_{n+1} \cdot \frac{1}{a_n}|=|\frac{(-3i)^{n+1}z^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3i)^nz^{n}}| =|(-3i)z \frac{n^3}{(n+1)^3}| \\=|-3iz| \cdot | (\frac{n}{n+1})^3|=|-1||3||iz| |(1)^3| \\ =1(3)|z|(1)=3|z| \\ \text{ so we have } |\frac{a_{n+1}}{a_n}|=3|z|<1 \\ \text{ solving for } |z| \text{ gives } |z|<\frac{1}{3} \text{ so radius is } \frac{1}{3}\]
\[|a_n|^\frac{1}{n}<1 \\ |(\frac{-2 i n+1}{3n-2i})^nz^n|^\frac{1}{n}<1 \\ |\frac{-2i n+1}{3n-2i}z|<1 \\ |\frac{-2i n}{3n}z|<1 \\ |-1| |\frac{2}{3} |iz|<1 \\ \frac{2}{3}|z|<1 \\ |z|<\frac{3}{2} \\ \text{ which says the radius is } \frac{3}{2}\]
I have no idea if this is what your teacher's way is or the book's
but both seem to get me what wolfram also says as the answer

- Loser66

For the second one
1) why do you let it <1 ?
2) How can you get \( from~~ |\frac{-2i n+1}{3n-2i}z|<1 \\to~~ |\frac{-2i n}{3n}z|<1 \)

- freckles

1) http://tutorial.math.lamar.edu/Classes/CalcII/RootTest.aspx just using root test
2) n goes to infinity (you have the same degree on top and bottom; just take coefficients on top and bottom of term with highest exponent)

- freckles

wait was 1) for both questions I did or for the second one only?

- freckles

because I'm just using the standard root and ratio test

- freckles

http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

- Loser66

oh, you use root test!! not limsup as what the book says. Both them use |a_n|^(1/n) , and that confused me.

- Loser66

But I still not understand how you simplify it nicely like that, please, put some calculation to help me understand it better.

- freckles

which part
are you still not convinced |iz|=|z| ?

- Loser66

oh!! got it, you take limit so that it counts only the first numbers on both numerator and denominator. (the second one) (-2in / 3n)

- freckles

right \[\lim_{n \rightarrow \infty} \frac{\color{red}{-2i } n +1}{\color{red}3n-2i}=\frac{-2i}{3}\]

- Loser66

|dw:1444011753965:dw|

- freckles

yeah I did put a disclaimer above all of my work

- freckles

"we are doing n goes to infinity
so I will probably not write the word limit...
"

- Loser66

Thank you so much. I do appreciate.

- freckles

lol sorry I was feeling lazy didn't feel like writing it over and over

- freckles

anyways I have to go peace

- Loser66

Have a good night.

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