Imaginary number.... ???

[penguin0708]

Hi-

I'm new to the site. : )

I'm trying to figure out something that's on a practice test. I'm actually trying to learn it, not just figure out the answer.

The problem is:
The imaginary number i is defined such that i<squared> = -1. What does i + i<squared> + i<cubed> + ... + i<23rd> equal?

The possible answers are:
i
-i
-1
0
1

I really appreciate any help on this!!!

-penguin0708

 

[Deleted member 4993]

Remember - series problems?

You are asked to find the sum of a particular geometric series.

 

[penguin0708]

Hi-

Okaaay... That sorta helps. I looked up the series problems thing. But I still don't get the whole imaginary number thing. Sorry, I'm not a math person. I'm just trying to learn the concept so I can pass this test and skip 3 college math classes. If you could sorta walk me through it, that would be really awesome.

Thanks so much for any help!!!
-penguin0708

 

[Deleted member 4993]

Hi-

Okaaay... That sorta helps. I looked up the series problems thing. But I still don't get the whole imaginary number thing. Sorry, I'm not a math person. I'm just trying to learn the concept so I can pass this test and skip 3 college math classes. If you could sorta walk me through it, that would be really awesome.

Thanks so much for any help!!!
-penguin0708

So what is the formula for summation of a geometric series?

 

[pka]

\(\displaystyle \L \begin{array}{rcl}
S & = & \sum\limits_{k = 1}^{23} {i^k } \\
iS & = & \sum\limits_{k = 1}^{23} {i^{k + 1} } = \sum\limits_{k = 2}^{24} {i^k } \\
iS - S & = & i^{24} - i \\
S & = & \frac{{1 - i}}{{i - 1}} \\
\end{array}\)

 

[penguin0708]

Hi-

Okay, thanks. I still don't get it, but now I know that I don't know that. :lol: I think I can pass the test without that, I just was making sure it wasn't something super-simple that I could do some review work on and have a better chance of passing the test.

Thanks everyone so much for your help!!
I will definitely use this site later when I have horrible homework. This is a great site!!!
-penguin0708 :wink:

 

[Deleted member 4993]

Hi-

I just was making sure it wasn't something super-simple ....

It is really very simple - generally covered in high school algebra

\(\displaystyle Sum = S = a + ar + ar^2 + ar^3 ......ar^n..................(1)\)

\(\displaystyle Sr = ar + ar^2 + ar^3 ......ar^{n+1}...............................(2)\)
Subtract (1) from (2)
\(\displaystyle Sr - S = ar ^{n+1} - a\)

\(\displaystyle S(r - 1) = a ( r^{n+1} - 1}\)

\(\displaystyle S = a \frac{r^{n+1} - 1}{r-1} \\)

In your case

a = i

r = i

n = 22

and \(\displaystyle i^{23} = -i\)

In a college level math class (no matter for what major) - you should not have difficulty in following this. If you are still uncomfortable, I think you should review your high-school math rigorously.

 

[soroban]

Hello, penguin0708!

Welcome aboard!

Here's another approach . . .

The imaginary number \(\displaystyle i\) is defined such that: \(\displaystyle \,i^2 \:=\:-1\)

What does: \(\displaystyle \:i\,+\,i^2\,+\,i^3\,+\,\cdots\,+\,i^{23}\) equal?

The possible answers are: \(\displaystyle \;-i\;\;\;-1\;\;\;0\;\;\;1\)

Note the first four powers of \(\displaystyle i:\;\;\begin{array}{ccc}i & = & i \\ i^2 & = & -1 \\ i^3 & = & -i \\ i^4 & = & 1\end{array}\)

Hence, the sum of four consecutive powers is zero:
. . \(\displaystyle i\,+\,(-1)\,+\,(-i)\,+\,(1) \;=\;0\)

Hence, 23 terms of the series gives us: \(\displaystyle \:0\,+\,0\,+\,0\,+\,0\,+\,0\,+\,\left[ i\,+\, (-1)\,+\, (-i)\right] \;=\;\fbox{-1}\)