Question

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

Answer 1

@oldrin.bataku

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

I am sorry, my computer is lagging.

Answer 7

I don't have the answers for them.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

it seems the other way you have it also works

Answer 15

I show you the book.

Answer 16

theorem 1.3 and proposition 1.4

Answer 17

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 \(lim|\dfrac{a_n}{a_{n+1}}|= lim\frac|{(-3i)^n}{n^3}*\dfrac{(n+1)^3}{(-3i)^{n+1}}|=|\dfrac{1}{(-3i)}lim\dfrac{n^3}{(n+1)^3}=\dfrac{1}{|-3i|}\)

Answer 18

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

Answer 19

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

Answer 20

They are completely different , right?

Answer 21

|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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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)

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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)

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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