Imaginary Number Problem: Finding the Sum of Powers of i - March 12, 2012

[Jameson]

The imaginary number, $i$ is defined such that $i^2=-1$. What does $i+i^2+i^3+...i^{23}$ equal? Explain your reasoning.
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) soroban
3) veronica1999
4) anemone
5) mathmaniac
6) BAdhi

This week we received solutions with very different approaches to solving the problem so I will include more solutions than usual.

Solution:

(from MarkFL)

Spoiler

Since $i^n=i^{n\pm4m}$ with $n,m\in\mathbb{Z}$ we may write:

$\displaystyle \sum_{k=1}^{23}i^k=5\left(i+i^2+i^3+i^4 \right)+\left(i+i^2+i^3 \right)=5\left(i-1-i+1 \right)+\left(i-1-i \right)=5(0)-1=-1$

(from soroban)

Spoiler

Let $S \,=\,i + i^2 + i^3 + \cdots + i^{23}$

Then $S+1 \:=\:1 + i + i^2 + i^3 + \cdots + i^{23}$

The right side is a geometric series
. . [/color]with first term $a = 1$, common ratio $r = i$, and $n = 24$ terms.

Its sum is: .[/color]$1\cdot\dfrac{1 - i^{24}}{1-i} \:=\:\dfrac{1-1}{1-i} \:=\:0$Therefore: .[/color]$S+1 \;=\;0 \quad\Rightarrow\quad S \:=\:-1$

(from anemone)

Spoiler

$\displaystyle i+i^2+i^3+i^4+i^5+i^6+i^7+i^8+\cdots+i^{22}+i^{23}$

$\displaystyle=i+i^2+i^2.i+(i^2)^2+(i^2)^2.i+(i^2)^3+(i^2)^3.i+(i^2)^4+(i^2)^4.i+\cdots++(i^2)^{11}+(i^2)^{11}.i$

$\displaystyle=i-1-i+1+i+-1-i+1+i-\cdots-i-1$

Now, if we group the second and third terms together and fourth and fifth terms together, and so on and so forth, i.e.

$\displaystyle=i-(1+i)+(1+i)-(1+i)+(1+i)-\cdots-(i+1)$

we see that the first group and second group cancel out perfectly and this pattern continues for another total of 4 pairs of such cancel-able terms and yields the following result:

$\displaystyle=i-\cancel {(1+i)}+\cancel {(1+i)}-\cancel {(1+i)}+\cancel {(1+i)}-\cdots-\cancel {(1+i)}+\cancel {(1+i)}-(i+1)$

$\displaystyle=i-(1+i)$

$\displaystyle=-1$

(from BAdhi)

Spoiler

$$\begin{align*}

P&=\sum \limits_{k=1}^{23} i\\
&=i + \sum \limits_{k=1}^{11}i^{2k}+i^{2k+1}\\
&=i+ \sum \limits_{k=1}^{11}i^{2k}(1+i)\\
&=i+ \sum \limits_{k=1}^{11}(i^2)^k(1+i)\\
&=i+ \sum \limits_{k=1}^{11}(-1)^k\underbrace{(1+i)}_{a}\\
&=i+ \sum \limits_{k=1}^{11}(-1)^ka\\
&=i+(-1)a+\sum \limits_{k=2}^{11}(-1)^ka\\
&=i-(1+i)+\sum \limits_{k=1}^{5}(-1)^{2k}a+(-1)^{2k+1}a\\
&=-1+\sum \limits_{k=1}^{5}\underbrace{a+(-1)a}_{=0}\\
&=-1\\

\end{align*}$$
therefore,
$$i+i^2+i^3+\cdots+i^{23}=-1$$