Four answers, endlessly repeating.
Simplify
i975
No calculator can do this one.
Powers of i cycle every four: i, −1, −i, 1. Divide 975 by 4 and the remainder is 3, so i⁹⁷⁵ = i³ = −i. i is the trap: it divides only the last digit, 5 ÷ 4 = 1 r 1 — but 10 is not a multiple of 4, so the remainder needs the last two digits, 75. Only the remainder matters, which is exactly why no calculator is required.
Everything rests on one fact: i squared is −1.
The fifth power is i again, and the sixth is −1.
Because the loop closes after four steps, only the leftover after dividing the exponent by 4 matters.
You only need the remainder, never the quotient.
975 ÷ 4 = 243 remainder 3
we ignore these
this is the answer
Faster way: only the last two digits matter.
975 → 75 ÷ 4 = 18 remainder 3
100 is divisible by 4, so the hundreds never change the remainder.
So 4 · 243 = 972, and 975 is three steps past it.
i975: last digit 5, 5 ÷ 4 = 1 r 1, so “i” ✗. That shortcut works for divisibility by 2 and 5, not 4 — 10 leaves remainder 2 when divided by 4. Keep the last two digits: 75 ÷ 4 = 18 r 3, so −i.
| Remainder | Same as | Value |
|---|---|---|
| 1 | i1 | i |
| 2 | i2 | −1 |
| 3 | i3 | −i |
| 0 | i4 | 1 |
Remainder 3, so i975 behaves like i3.
Remainder 0 means a full loop, which lands on 1.
Break the exponent into a multiple of 4 plus the leftover.
i975 = i972 · i3
972 + 3 = 975
i972 = (i4)243
972 = 4 · 243
(i4)243 = 1243 = 1
any power of 1 is 1
i975 = 1 · i3 = −i
All the full loops collapse to 1 and disappear. Only the leftover steps are left doing anything.
The same law runs the other way for a product: multiply by adding the exponents first. i³⁷ × i⁷ = i⁴⁴, and 44 ÷ 4 leaves remainder 0, so the product is 1.
Negative exponents: i⁻¹ and i⁻⁷.
i⁻¹ = 1i = ii² = i−1 = −i
multiply top and bottom by i
i⁻⁷ = i⁻⁷ · i⁸ = i¹
i⁸ = 1, so adding 8 changes nothing
= i
check: 1i⁷ = 1−i = i ✓
What is different: we cannot take a remainder of −7 the usual way. Add a multiple of 4 until the exponent is positive.
A letter in the exponent: i4n + 6, then i2n, for a positive integer n.
i4n + 6 = (i⁴)ⁿ · i⁶ = i⁶
4n is n full loops
6 ÷ 4 = 1 r 2 → −1
whatever n is
i2n = (i²)ⁿ = (−1)ⁿ
1 if n is even, −1 if n is odd
What is different: 4n is a whole number of loops, but 2n is not always — so i2n has no single value.
A coefficient inside the bracket: (2i)⁵.
(2i)⁵ = 2⁵ · i⁵
the power takes both factors
= 32 · i = 32i
5 ÷ 4 = 1 r 1
What is different: the 2 is raised as well. Writing 2i⁵ = 2i forgets that the bracket holds a product.
Find i + i² + i³ + … + i43.
i + (−1) + (−i) + 1 = 0
every block of four cancels
43 terms = 10 blocks (40 terms) + 3 left
43 ÷ 4 = 10 r 3
i41 + i42 + i43 = i − 1 − i = −1
only the leftovers survive
Find (1 − i)6.
(1 − i)² = 1 − 2i + i² = −2i
square once — the 1 and −1 cancel
(1 − i)6 = (−2i)³ = −8i³
cube the square
= −8(−i) = 8i
i³ = −i
The sum keeps the leftover terms, so add them — do not just report the value of the last power, −i. And (1 − i)² is not 1 + i² = 0: the middle term −2i is the whole answer. Similarly (1 + i)² = 2i.
Expect a single huge power, a sum or difference of two, a long run i + i² + … , or a power of (1 ± i). Reduce every exponent by its remainder on division by 4, cancel full blocks of four, and square (1 ± i) first.
Count it out: i, −1, −i, 1, i, −1, −i
The 3rd and the 7th both land on −i.
Add two powers
Simplify
i2026 + i2027
2026 ÷ 4 leaves remainder 2, so i²⁰²⁶ = i² = −1. 2027 leaves remainder 3, so i²⁰²⁷ = i³ = −i. Adding gives −1 − i. C uses i³ = +i, but i³ = i² · i = −i. Take each remainder separately before combining.
Hint: work out each power separately, then add. The answer will have a real part and an imaginary part.
