Expert Answer to "Adding powers of i Ive"

Answered question

2022-01-17

Adding powers of i
Ive

nick1337

Expert2022-01-19Added 777 answers

Step 1 Observing that

i^{3} + i^{4} + \dots + i^{50}

is a geometric progression with ratio i, first term

i^{3}

and

50 - 3 + 1 = 48

terms, we have

i^{3} + i^{4} + \dots + i^{50} = i^{3} \times \frac{1 - i^{50 - 3 + 1}}{1 - i}

= i^{3} \times \frac{1 - i^{48}}{1 - i}

i^{2} i \times \frac{1 - (i^{2})^{24}}{1 - i}

= - i \frac{1 - (- 1)^{24}}{1 - i} = - i \frac{1 - 1}{1 - i} = 0

Edit: arithmetic corrected to geometric

Vasquez

Expert2022-01-19Added 669 answers

From

i^{2} = - 1

you get

i^{4 n} = 1

i^{4 n + 1} = i

i^{4 n + 2} = - 1

and

i^{4 n + 3} = - i

Then you just count your positive and negative multiples of 1 and i. In particular,

i^{3} + i^{4} + \dots + i^{50} = 0

alenahelenash

Expert2022-01-24Added 556 answers

Hint

i^{3} + i^{4} + \dots + i^{k} = 0 \Leftrightarrow k \equiv 2

(mod 4) Generally, suppose that z has order

m > 1

. Therefore

z^{n} = 1 \Leftrightarrow m | n

hence: Lemma

z^{j} + z^{j + 1} + \dots + z^{k} = 0 \Leftrightarrow k \equiv j - 1

(mod m) Proof:

z^{j} (1 + z + \dots + z^{k - j}) = z^{j} \frac{1 - z^{k - j + 1}}{1 - z} = 0 \Leftrightarrow m | k - j + 1

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