Thread: [SOLVED] Convergence of a series

1. [SOLVED] Convergence of a series

I have a question that I can't do...can anybody help me?

Investigate the convergence of the series (in the complex plane) for difference values of a and b:

Generalise your solution by determining what constraints are required on the values of a and b, so as to guarantee the convergence of the series?

$\displaystyle S(a,b) = \sum_{n=0}^{\infty}(a+bi)^n$

Thanks if you can help.

2. $\displaystyle a^2 + b^2 < 1$ assuminf a and b are both real

3. Sorry, still trying to understand

But I thought a + bi is a complex number.
i.e. b is not real.

4. Originally Posted by Bucephalus
But I thought a + bi is a complex number.
i.e. b is not real.
in the form a + bi, both a and b are real. that is generally how complex numbers are defined. we get a complex number because there is an "i" attached to the real number b

5. so b could be a complex number?

So the assumption was that a and b are real numbers i.e.
it is not of the form a + (c +di)i
that's what I now understand of that that assumption is assuming.
Is this correct?

6. Originally Posted by Bucephalus
So the assumption was that a and b are real numbers i.e.
it is not of the form a + (c +di)i
that's what I now understand of that that assumption is assuming.
Is this correct?
correct. a and b are considered to be real unless otherwise stated.

and it would not be otherwise stated, since even a + (c + di)i can be rewritten in the form a + bi

7. well why isn't it....

$\displaystyle \sqrt{a^2 + b^2} < 1$
Because I thought that the condition for geometric series convergence is

|r| < 1 for geomtric series

$\displaystyle S = \sum_{n=0}^{\infty}ar^n$

thanks.

8. Originally Posted by Bucephalus
$\displaystyle \sqrt{a^2 + b^2} < 1$
Because I thought that the condition for geometric series convergence is

|r| < 1 for geomtric series

$\displaystyle S = \sum_{n=0}^{\infty}ar^n$

thanks.

Thanks.