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## Loser66 one year ago 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

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1. Loser66

@oldrin.bataku

2. freckles

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

3. 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

4. freckles

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

5. freckles

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

6. Loser66

I am sorry, my computer is lagging.

7. Loser66

I don't have the answers for them.

8. 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

9. 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

10. Loser66

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

11. freckles

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

12. Loser66

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

13. 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

14. freckles

it seems the other way you have it also works

15. Loser66

I show you the book.

16. Loser66

theorem 1.3 and proposition 1.4

17. Loser66

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

19. freckles

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

20. Loser66

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

21. Loser66

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

22. 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

23. Loser66

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

24. 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

25. Loser66

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

26. freckles

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

27. 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|

28. Loser66

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

29. Loser66

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

30. 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

31. Loser66

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

32. 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)

33. freckles

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

34. freckles

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

35. freckles
36. Loser66

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

37. Loser66

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

38. freckles

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

39. 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)

40. freckles

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

41. Loser66

|dw:1444011753965:dw|

42. freckles

yeah I did put a disclaimer above all of my work

43. freckles

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

44. Loser66

Thank you so much. I do appreciate.

45. freckles

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

46. freckles

anyways I have to go peace

47. Loser66

Have a good night.

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