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How do I do this... 2i^19 - 5i^40 + 4i^97 + 9i^122?
The directions say to simplify.
8 Answers
- Iggy RockoLv 74 years ago
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
See the pattern?
2i^19 - 5i^40 + 4i^97 + 9i^122 =
2i^3 - 5i^4 + 4i^3 + 9i^2 =
-2i - 5 - 4i - 9 =
-14 - 6i
- ?Lv 74 years ago
i⁰ = 1+
i¹ = i
i² = -1
i³ = -i
i⁴ = 1
so take these exponents mod 4:
2i^19 - 5i^40 + 4i^97 + 9i^122
= 2i³-5+4i+9i²
= -2i-5+4i-9
= -14+2i
- ?Lv 74 years ago
i^2 = −1
i^4 = 1
2i^19 − 5i^40 + 4i^97 + 9i^122
= 2(i^4)^4(i^2)(i) − 5(i^4)^10 + 4(i^4)^24(i) + 9(i^4)^30(i^2)
= 2(1)(−1)(i) − 5(1) + 4(1)(i) + 9(1)(−1)
= −2i − 5 + 4i − 9
= −14 + 2i
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Alternate method:
i^1 = i
i^2 = −1
i^3 = −i
i^4 = 1
i^5 = i
i^6 = −1
i^7 = −i
i^8 = 1
and the pattern continues...
i^(4k) = 1
i^(4k+1) = i
i^(4k+2) = −1
i^(4k+3) = −i
2i^19 − 5i^40 + 4i^97 + 9i^122
= 2i^(4*4+3) − 5i^(4*10) + 4i^(4*24+1) + 9i^(4*30+2)
= 2(−i) − 5(1) + 4(i) + 9(−1)
= −2i − 5 + 4i − 9
= −14 + 2i
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- 4 years ago
i = √-1
i^2 = -1
i^3 = -√-1 = -i
i^4 = 1
And the patter repeats. So i^19 is the same as i^3 (since 19/4 has a remainder of 3). The same reasoning applies to the next terms also.
2i^19 - 5i^40 + 4i^97 + 9i^122
2i^3 - 5i^4 + 4i^1 + 9i^2
2(-i) - 5(1) + 4(i) + 9(-1)
-2i - 5 + 4i -9
2i-14
- JLv 74 years ago
Reduce exponents mod 4, because i,i^2,i^3,i^4 cycle since i^4=1, so
2i^19 - 5i^40 + 4i^97 + 9i^122 =
2i^3 - 5i^0 + 4i^1 + 9i^2 =
-2i - 5 + 4i - 9 =
2i - 14
- ?Lv 74 years ago
The powers of i cycle every 4 powers; i.e., i = sqrt(-1), i^2 = -1, i^3 = -i, and i^4 = 1. Divide that exponent of i by 4; whatever the remainder is will correspond to one of the aforementioned powers of i up to i^4 = 1, which results if the exponent is divisible by 4. So:
2i^19 - 5i^40 + 4i^97 + 9i^122
= 2(i^3) - 5(i^4) + 4(i^1) + 9(i^2)
= 2(-i) - 5(1) + 4(i) + 9(-1)
= -2i - 5 + 4i - 9
= -14 + 2i. Final.
- ?Lv 74 years ago
i^19 = i^18*i = (i^2)^9 * 1 = (-1)^9 * i = - i so the first term reduces to 2*(-1) = -2. i^40 = (i^2)^20 = (-1)^20 = + 1 so the second term is 5*1 = + 5; i^122 = (i^2)^61 = (-1)^61 = -1 so the third term is 9*(-1) = -9 so altogether you have -2 + 5 - 9 = -11 + 5 = -6
Source(s): Retired Maths Teacher