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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 $i+i^2+i^3+\cdots+i^{23}$?

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marked as duplicate by Rasmus, Henry T. Horton, Asaf Karagila, QiL'8, Javier Álvarez Feb 11 '13 at 22:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 1:42
Work $\pmod 4$. – Pedro Tamaroff Feb 10 '13 at 1:58
up vote 6 down vote accepted

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


  • $i + i^2 + i^3 + i^4\;\; = \;\;i + -1 + - i + 1 \;\;=\;\; 0$

  • 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 $i^n = i^k$

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

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).

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This works, but requires dividing complex numbers. Given the question, that may not be available. – Ross Millikan Feb 10 '13 at 2:18
@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?" – Arjang Feb 10 '13 at 2:20
@RossMillikan: Actually, it doesn't really require dividing by complex numbers... – trb456 Feb 10 '13 at 2:28
@trb456: the denominator is $1-i$ – Ross Millikan Feb 10 '13 at 2:31
@RossMillikan: Right, but the numerator is... – trb456 Feb 10 '13 at 2:33

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

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

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

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Also note that $i + i^2 + i^3 + i^4 = i^5 + i^6 + i^7 + i^8 = \cdots$ – Parth Kohli Feb 10 '13 at 15:30
@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

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 $-i$.

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

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

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