# Geometric Sum, involving complex numbers

• Jun 3rd 2011, 03:59 PM
Zellator
Geometric Sum, involving complex numbers
Hi Forum!

I came across an exercise involving both geometric sum and complex numbers.

I've read through Wikipedia and found: The summation formula for geometric series remains valid even when the common ratio is a complex number.

So, for the question:
A sequence has ten terms with first term $\displaystyle i$, for$\displaystyle i = \sqrt-1$. Each subsequent term is 2i times its previous term. That is, the 2nd term is $\displaystyle i.2i$ .
The sum of this sequence is a + bi . What is the value of |a + b|?

If we wish to get the sum of the geometric progression we use:
$\displaystyle S_n=a(1-r^n)/(1-r)$

Using this you only give headaches, since we are searching for a+bi solution.
So, what would be the best approach? --besides from calculating all the terms up to n=10.

Is it really possible to use the geometric sum formula here?

Thanks!
• Jun 3rd 2011, 04:28 PM
mr fantastic
Quote:

Originally Posted by Zellator
Hi Forum!

I came across an exercise involving both geometric sum and complex numbers.

I've read through Wikipedia and found: The summation formula for geometric series remains valid even when the common ratio is a complex number.

So, for the question:
A sequence has ten terms with first term $\displaystyle i$, for$\displaystyle i = \sqrt-1$. Each subsequent term is 2i times its previous term. That is, the 2nd term is $\displaystyle i.2i$ .
The sum of this sequence is a + bi . What is the value of |a + b|?

If we wish to get the sum of the geometric progression we use:
$\displaystyle S_n=a(1-r^n)/(1-r)$

Using this you only give headaches, since we are searching for a+bi solution.
So, what would be the best approach? --besides from calculating all the terms up to n=10.

Is it really possible to use the geometric sum formula here?

Thanks!

I don't see what would cause a "headache". Roll up your sleeves, use the formula, get an answer and then convert the answer into the form a + ib.

If you need more help, please show your work and say where you are stuck.
• Jun 3rd 2011, 05:25 PM
Zellator
Quote:

Originally Posted by mr fantastic
I don't see what would cause a "headache". Roll up your sleeves, use the formula, get an answer and then convert the answer into the form a + ib.

If you need more help, please show your work and say where you are stuck.

Hi mr fantastic (sorry about the wrong section thing, the test is said to be of the lowest level of 3, I thought it was alright.)

Ok, this is my first contact with complex numbers so:

$\displaystyle a_1_0=-2^1^0$ since $\displaystyle i.(2i)^9=2^9.i^1^0$
$\displaystyle a_1=i$
$\displaystyle r=2i$
$\displaystyle S_1_0=[i-2i(-2^9)]/(1-2i)$
$\displaystyle S_1_0=(i+2^1^0i)/(1-2i)$
$\displaystyle S_1_0=i(1+2^1^0)/(1-2i)$

We could go a little further and multiply the whole equation by (1-2i), but this would give the first equation -gradually- once more.
Am I missing something here?

Thanks. :)
• Jun 3rd 2011, 06:11 PM
jgv115
Are you trying to multiply by the conjugate? If you are you are doing it wrong because the conjugate is (1+2i) not (1-2i)

Also I'm not quite sure how you got your numerator...

The formula says $\displaystyle a(1-r^n)$

We know n is 10, a is i and r is 2i.

$\displaystyle i(1-(2i)^{10})$

Have you learnt to deal with complex numbers...?

You have to know that $\displaystyle i^2 = -1$ so $\displaystyle i^3 = -i$
• Jun 4th 2011, 02:51 AM
Zellator
Quote:

Originally Posted by jgv115
Are you trying to multiply by the conjugate? If you are you are doing it wrong because the conjugate is (1+2i) not (1-2i)

Also I'm not quite sure how you got your numerator...

The formula says $\displaystyle a(1-r^n)$

We know n is 10, a is i and r is 2i.

$\displaystyle i(1-(2i)^{10})$

Have you learnt to deal with complex numbers...?

You have to know that $\displaystyle i^2 = -1$ so $\displaystyle i^3 = -i$

Hi jgv115!
Yes, I know that $\displaystyle i^3 = -i$, that's why we get $\displaystyle i+2^1^0i$.