What is the value of $i+i^2+i^3+\cdots+i^{23}$?

Question

Can anyone help me with this question and show me a step by step solution please?

The imaginary number is $i$ is defined such that $i^2=-1$. What is $i+i^2+i^3+\cdots+i^{23}$?

To get multi-character exponents, enclose them in braces, so use i^{23} to get $i^{23}$ The same thing works throughout $\LaTeX$, whatever is in braces acts like one character. — Ross Millikan, Feb 10 '13 at 01:42

score 12 · Answer 1 · answered Feb 10 '13 at 02:17

12

Just use the geometric series :

$$\sum\limits_{k = 0}^{n-1} {r^{k} = \frac{1-r^n}{{1 - r}}}$$

Set $r= i , n=24$ and remember to take away 1 ( your summation starts from k=1 , the above formula starts from k=0).

answered Feb 10 '13 at 02:17

jimjim

9,855

2

This works, but requires dividing complex numbers. Given the question, that may not be available. – Ross Millikan Feb 10 '13 at 02:18
@RossMillikan : fair enough! but I like to throw in the math equivalent of monkey wrench into the works! Make them ask "why we can't use this?" – jimjim Feb 10 '13 at 02:20
1

@RossMillikan: Actually, it doesn't really require dividing by complex numbers... – Feb 10 '13 at 02:28
@trb456: the denominator is $1-i$ – Ross Millikan Feb 10 '13 at 02:31
3

@RossMillikan: Right, but the numerator is... – Feb 10 '13 at 02:33

score 11 · Answer 2 · answered Feb 10 '13 at 01:41

11

Hint: do you know what $i^2, i^3,$ and $i^4$ are? That should lead you to expect a pattern.

answered Feb 10 '13 at 01:41

Ross Millikan

383,099

what does i to the power of 1 equal? – Jon Feb 10 '13 at 02:00
Little Jon: $i^1 = i$, just as $2^1 = 2$, and $i^0 = 1$, just like $2^0 = 1.$ – amWhy Feb 10 '13 at 02:01

score 11 · Answer 3 · answered Feb 10 '13 at 02:15

11

Hint: Since $i^2=-1$, we have $\color{#C00000}{i}+\color{#00A000}{i^2}+\color{#C00000}{i^3}+\color{#00A000}{i^4}=0$.

answered Feb 10 '13 at 02:15

robjohn

353,833

Also note that $i + i^2 + i^3 + i^4 = i^5 + i^6 + i^7 + i^8 = \cdots$ – P.K. Feb 10 '13 at 15:30
1

@Novice: we even have $i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}=i^n(i^1+i^2+i^3+i^4)=i^n\cdot0$ – robjohn Feb 10 '13 at 19:08

amWhy · Accepted Answer · 2013-02-11T04:01:07.060

6

Look at the values of $i^0,\, i, \,i^2, \,i^3,\,i^4,\,$...$,\,i\,^5,...$: you'll see a cyclic pattern emerge.

Observations:

$i + i^2 + i^3 + i^4\;\; = \;\;i + -1 + - i + 1 \;\;=\;\; 0$
For any $n\in \mathbb{N}$, $i^n = i,\, i^2,\, i^3,\, i^0 = i^4$.
That is, for any given $n,\,$ consider $\;k\; = \;n\pmod 4.\;$ Then $i^n = i^k$

edited Feb 11 '13 at 04:01

answered Feb 10 '13 at 01:47

amWhy

210,739

I understand the pattern now. Thank you made it simple. – Jon Feb 10 '13 at 17:50
Great! You're welcome, Little Jon! – amWhy Feb 10 '13 at 17:52

score 4 · Answer 5 · answered Feb 10 '13 at 03:05

4

Hint $ $ Call it $\,S.\,$ $\, (1\!-\!i)(1\!+\!S) = (1\!-i)(1\!+i\!+\cdots\! + i^{23}) = 1\!-i^{24} = 1\!-(-1)^{12}\! = 0\:$ $\Rightarrow$ $\,S = -1.$

answered Feb 10 '13 at 03:05

Math Gems

20,090

score 4 · Answer 6 · answered Feb 10 '13 at 04:53

Multiplication of complex numbers is, geometrically, rotation.

The angle between the positive real number line and a given complex number is called the argument of that number, usually given in radians. When two complex numbers are multipled together, the product's argument is the sum of the arguments of the two factors.

The Euclidean distance of a complex number from $0$ is its modulus. When two complex numbers are multiplied together, the modulus of the product is the product of the two moduli.

Because the argument of $i$ is $90$ degrees ($\pi$ radians), and its modulus is $1$, when we multiply a number by $i$, we rotate that number by 90 degrees.

So $i$ is the unit vector pointing straight up. Then $i\times i = i^2$ is a vector pointing straight left ($-1$), $i^3$ is a vector pointing down ($-i$), and $i^4$ is a unit vector pointing right ($1$). If we add these together, we get zero because they all cancel out.

The subsequent powers just keep going around the unit circle in the same pattern: $90, 180, 270, 0, 90$ ... or $i, -1, -i, 1, i, ...$.

You have $23$ powers being added. The first $20$ are just five groups of four which each cancel to zero individually, and so add to zero overall. You're left with three powers which must continue around the circle like this: $i, -1, -i$. The sum of these is $-1$, since $i$ cancels $-i$.

score 1 · Answer 7 · answered Feb 10 '13 at 02:20

1

$$i\cdot \frac{i^{23}-1}{i-1}=\frac{i}{i-1} \cdot (i^{23}-1)=\frac{i(i+1)}{-2} \cdot (-i-1)=\frac{i(i+1)^2}{2}=\frac{i(1+2i-1)}{2}=-1.$$

answered Feb 10 '13 at 02:20

Iuli

6,870

What is the value of $i+i^2+i^3+\cdots+i^{23}$?

7 Answers7