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

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• Adding powers of $i$ 4 answers

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

Answer 1

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 $i^n = i^k$

Answer 2

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).

Answer 3

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

Answer 4

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

Answer 5

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 $-i$.

Answer 6

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

Answer 7

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