Please help me solve--
i18 + i16 + i14 + 1=
Thank you!!
2007-05-05 03:44:50 · 5 answers · asked by kk_junebug 2 in Science & Mathematics ➔ Mathematics
Please help me solve--
i18 + i16 + i14 + 1 =
My options are: i 1 0 -1
I can get to 48i +1 but where from there?
Thank you!!
2007-05-05 03:57:04 · update #1
Sorry - I need to convert this to a + bi form. Thanks!! :)
2007-05-05 03:58:39 · update #2
i18 + i16 + i14 + 1=
I will presume you mean
i^18 + i^16 + i^14, and that i is √(-1). This site is terrible for math notation.
First i^2 = -1
Second, (-1)^2 = 1
Third (a^x)(a^y) a^(x+y)
Fourth (a^x)^y = a^(xy)
With those tidbits in mind, taking each term separately...
i^18 = (i^2)^9 = (-1)^9 = (-1)(-1)^8 = (-1)((-1)^2)^4 = (-1)(1^4) = (-1)(1) = -1
i^16 = (i^2)^8 = (-1)^8 = ((-1)^2)^4) = 1^4 =1
i^14 = (i^2)^7 = (-1)^7 = (-1)(-1)^6 = (-1)((-1)^2)^3) = (-1)(1^3) = (-1)(1) = -1
Adding these three values we get:
i^18 + i^16 + i^14 = -1 + 1 + (-1) = -1
2007-05-05 04:36:01 · answer #1 · answered by gugliamo00 7 · 0⤊ 0⤋
= i(18 + 16 + 14) + 1 = 48i + 1
2007-05-05 03:49:32 · answer #2 · answered by kellenraid 6 · 0⤊ 0⤋
i(18+16+14)+1=1+48i
2007-05-05 03:50:26 · answer #3 · answered by Scott H 3 · 0⤊ 0⤋
The answer is
i38+1
because you cannot had the one with the rest of the imaginary numbers because the don't exist. So you have to add all of the imaginary (i) numbers then just tack on the 1.
You could almost consider the i as a variable.
2007-05-05 04:09:59 · answer #4 · answered by April E 2 · 0⤊ 0⤋
= 1 + 48 i
Will be an acceptable answer as it is in "a + b i" form.
2007-05-05 08:14:11 · answer #5 · answered by Como 7 · 0⤊ 0⤋