Sum of Imaginary numbers

Algebra Level 2

Evaluate the following sum:

i 0 + i 1 + i 2 + ⋯ + i 2020 i^0+i^1+i^2+\dots+i^{2020}

Notation: i = − 1 i = \sqrt{-1} denotes the imaginary unit .

i i 1 + i 1+i 0 0 1 − i 1-i 1 1

Hana Wehbi by Hana Wehbi , 51, USA

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3 solutions

David Vreken
Jul 25, 2020

i 4 n = 1 i^{4n} = 1 , i 4 n + 1 = i i^{4n + 1} = i , i 4 n + 2 = − 1 i^{4n + 2} = -1 , and i 4 n + 3 = − i i^{4n + 3} = -i for any integer n n .

Therefore:

i 0 + i 1 + i 2 + i 3 + i 4 + i 5 + i 6 + i 7 + . . . + i 2016 + i 2017 + i 2018 + i 2019 + i 2020 i^{0} + i^{1} + i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + ... + i^{2016} + i^{2017} + i^{2018} + i^{2019} + i^{2020}

= ( i 0 + i 1 + i 2 + i 3 ) + ( i 4 + i 5 + i 6 + i 7 ) + . . . + ( i 2016 + i 2017 + i 2018 + i 2019 ) + i 2020 = (i^{0} + i^{1} + i^{2} + i^{3}) + (i^{4} + i^{5} + i^{6} + i^{7}) + ... + (i^{2016} + i^{2017} + i^{2018} + i^{2019}) + i^{2020}

= ( 1 + i + ( − 1 ) + ( − i ) ) + ( 1 + i + ( − 1 ) + ( − i ) ) + . . . + ( 1 + i + ( − 1 ) + ( − i ) ) + 1 = (1 + i + (-1) + (-i)) + (1 + i + (-1) + (-i)) + ... + (1 + i + (-1) + (-i)) + 1

= ( 0 ) + ( 0 ) + . . . + ( 0 ) + 1 = (0) + (0) + ... + (0) + 1

= 1 = \boxed{1}

4 Helpful 1 Interesting 4 Brilliant 0 Confused

Thank you for sharing your solution, a brilliant method for solving it.

Hana Wehbi - 10 months, 3 weeks ago

Yeah, grouping is the method. I still made a mistake in that, unable to post a solution :( But it's same as yours so no problem! Still, I wonder whether there is any other method apart from grouping...

Mahdi Raza - 10 months, 2 weeks ago

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I was thinking to solve it as a geometric series! Definitely not going to work.

Hana Wehbi - 10 months, 2 weeks ago

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Yeah, cause we wouldn't be able to evaluate the denominator then ( i 0 ( i 2020 − 1 ) i − 1 ) \bigg( \dfrac{i^0(i^{2020} - 1)}{i-1}\bigg)

Mahdi Raza - 10 months, 2 weeks ago

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@Mahdi Raza – Or maybe we can, what if we rationalize the denominator

Mahdi Raza - 10 months, 2 weeks ago

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@Mahdi Raza – The top part is 0 still.

Hana Wehbi - 10 months, 2 weeks ago

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@Hana Wehbi – Hmmm, I didn't think of that... What's wrong though? Why isn't this method working, any idea?

Mahdi Raza - 10 months, 2 weeks ago

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@Mahdi Raza – That equation only works for real numbers − 1 < r < 1 -1 < r < 1

David Vreken - 10 months, 2 weeks ago

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@David Vreken – I agree but I saw a similar problem online where it was solved using geometric finite sum and it worked. I believe it has also to do with the power value.

Hana Wehbi - 10 months, 2 weeks ago

Here's another way to do it, but it's still grouping:

i 0 + i 1 + i 2 + i 3 + i 4 + i 5 + i 6 + i 7 + . . . + i 2016 + i 2017 + i 2018 + i 2019 + i 2020 i^0 + i^1 + i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + ... + i^{2016} + i^{2017} + i^{2018} + i^{2019} + i^{2020}

= ( i 0 + i 4 + . . . + i 2016 + i 2020 ) + ( i 1 + i 5 + . . . + i 2017 ) + ( i 2 + i 6 + . . . + i 2018 ) + ( i 3 + i 7 + . . . + i 2019 ) = (i^0 + i^4 + ... + i^{2016} + i^{2020}) + (i^1 + i^5 + ... + i^{2017}) + (i^2 + i^6 + ... + i^{2018}) + (i^3 + i^7 + ... + i^{2019})

= ( 506 ) + ( 505 i ) + ( − 505 ) + ( − 505 i ) = (506) + (505i) + (-505) + (-505i)

= 1 = \boxed{1}

I can't think of a different way.

David Vreken - 10 months, 2 weeks ago

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Thank you for sharing it.

Hana Wehbi - 10 months, 1 week ago

A different sub-approach, thanks!

Mahdi Raza - 10 months, 1 week ago
Chew-Seong Cheong
Jul 26, 2020

S n = i 0 + i 1 + i 2 + i 3 + i 4 + i 5 + i 6 + i 7 + ⋯ + i n = 1 + i − 1 − i ⏟ = 0 + 1 + i − 1 − i ⏟ = 0 + ⋯ + i n ⟹ S n = { 1 if n m o d 4 = 0 1 + i if n m o d 4 = 1 i if n m o d 4 = 2 0 if n m o d 4 = 3 \begin{aligned} S_n & = i^0 + i^1 + i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + \cdots + i^n \\ & = \underbrace{1 + i - 1 - i}_{=0} + \underbrace{1 + i - 1 - i}_{=0} + \cdots + i^n \\ \implies S_n & = \begin{cases} 1 & \text{if } n \bmod 4 = 0 \\ 1 + i & \text{if } n \bmod 4 = 1 \\ i & \text{if } n \bmod 4 = 2 \\ 0 & \text{if } n \bmod 4 = 3 \end{cases} \end{aligned}

Since 2020 m o d 4 = 0 2020 \bmod 4 = 0 , we have S 2020 = 1 S_{2020} = \boxed 1 .

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Thank you for sharing your solution.

Hana Wehbi - 10 months, 2 weeks ago
William Allen
Jul 29, 2020

i = − i 3 i = -i^3 and i 2 = − i 4 i^2 = -i^4 so that i + i 2 + i 3 + i 4 = 0 i+i^2+i^3+i^4 = 0 . Since 4 ∣ 2020 4|2020 every term apart from the first cancels and the answer is i 0 = 1 i^0=1 .

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Thank you for sharing your solution.

Hana Wehbi - 10 months, 2 weeks ago

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