Geometric Sum, involving complex numbers

**Zellator** · June 3rd 2011, 03:59 PM

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 $i$ , for $i = \sqrt-1$ . Each subsequent term is 2i times its previous term. That is, the 2nd term is $i.2i$ .
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:
$S_n=a(1-r^n)/(1-r)$

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!

**mr fantastic** · June 3rd 2011, 04:28 PM

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 $i$ , for $i = \sqrt-1$ . Each subsequent term is 2i times its previous term. That is, the 2nd term is $i.2i$ .
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:
$S_n=a(1-r^n)/(1-r)$

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.

**Zellator** · June 3rd 2011, 05:25 PM

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:

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

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.

**jgv115** · June 3rd 2011, 06:11 PM

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 $a(1-r^n)$

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

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

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

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

**Zellator** · June 4th 2011, 02:51 AM

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 $a(1-r^n)$

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

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

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

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

Hi jgv115!
Thanks for the reply!

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

So that was the problem; easy as it may seem, I've never learnt to deal with complex numbers.
That is a good thing to learn!
Ok, all sorted out here!

Thanks for your help.

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