Term numbers divisible by 4 equal 1. Notice that the first term is the zeroth term.
C
c
What number is 4n always divisible by, regardless of n?
D
d
i^(4n+1) describes the term that comes directly after i^(4n).
E
e
The number 396 is divisible by 4.
A
a
Table: See solution. Pattern: The pattern repeats 1, i, - 1, - i.
B
b
i^(16)=1
i^(25)=i
i^(39)=- i
i^(100)=1
C
c
1
D
d
i^(4n+1)=i
i^(4n+2)=- 1
i^(4n+3)=- i
E
e
i^(396)=1
i^(397)=i
i^(398)=- 1
i^(399)=- i
Practice makes perfect
a
We have already determined the values of i^0, i^1, and i^2. By factoring the next twelve terms so that one factor is i^2, we can determine its value by then substituting i^2=- 1 and simplifying.
c|l|l|c
Term & Factor & i^2=- 1 & Multiply
[-0.9em]
i^3 & (i^2)i & (- 1)i & - i [0.2em]
[-0.9em]
i^4 & (i^2)^2 & (- 1)^2 & 1 [0.2em]
[-0.9em]
i^5 & (i^2)^2i & (- 1)^2i & i [0.2em]
[-0.9em]
i^6 & (i^2)^3 & (- 1)^3 & - 1 [0.2em]
[-0.9em]
i^7 & (i^2)^3i & (- 1)^3i & - i [0.2em]
[-0.9em]
i^8 & (i^2)^4 & (- 1)^4 & 1 [0.2em]
[-0.9em]
i^9 & (i^2)^4 i & (- 1)^4 i & i [0.2em]
[-0.9em]
i^(10) & (i^2)^5 & (- 1)^5 & - 1 [0.2em]
[-0.9em]
i^(11) & (i^2)^5i & (- 1)^5i & - i [0.2em]
[-0.9em]
i^(12) & (i^2)^6 & (- 1)^6 & 1 [0.2em]
[-0.9em]
i^(13) & (i^2)^6i & (- 1)^6i & i [0.2em]
[-0.9em]
i^(14) & (i^2)^7 & (- 1)^7 & - 1 [0.2em]
[-0.9em]
i^(15) & (i^2)^7i & (- 1)^7i & - i
Having calculated all the terms, we can see a pattern that repeats itself in cycles of four.
1, i, - 1, - i^(1^(st) cycle), 1, i, - 1, - i^(2^(nd) cycle), 1, i, - 1, - i^(3^(rd) cycle), 1, i, - 1, - i^(4^(th) cycle)
The pattern repeats 1, i, - 1, and - i.
b
If we let the first number represent the zeroth term, n=0, we see that every fourth term is 1.
r|ccccc
n & 0^(th)& 1^(st)& 2^(nd)& 3^(rd) & 4^(th)
i^n & 1 & i & - 1 & - i & 1
Therefore, any term number that is divisible by 4 must have a corresponding value of 1. Since both 16, and 100 are numbers divisible by 4, we know that they must both be 1.
i^(16)&=1
i^(100)&=1
The expression i^(25) has a term number that is 1 above a multiple of 4. Similarly, i^(39) has a term number that is one below a multiple of 4. With this information, we can determine their respective values.
r|ccccc
n & 24^(th)& 25^(th)& 26^(th)& 27^(th) & 28^(th)
i^n & 1 & i & - 1 & - i & 1
r|ccccc
n & 36^(th) & 37^(th)& 38^(th)& 39^(th) & 40^(th)
i^n & 1 & i & - 1 & - i & 1
As we can see, i^(25)=i and i^(39)=- i.
c
From Part B, we determined that any term number in the sequence divisible by 4 must have a value of 1. If we multiply n by 4, we guarantee that whatever number we substitute for n will always be divisible by 4. c|c|c
n & 4n/4 & = [0.8em]
[-0.6em]
5 & 4( 5)/4 & 5 [1em]
[-0.6em]
20 & 4( 20)/4 & 20 [1em]
[-0.6em]
190 & 4( 198)/4 & 198
Therefore, we know that i^(4n)=1.
d
From Part C, we know that i^(4n)=1. Therefore, i^(4n+1) must describe the term that comes after i^(4n), while i^(4n+2) and i^(4n+3) describes the third and fourth term after i^(4n).
r|ccccc
n & 4n& 4n+1 & 4n+2& 4n+3
i^n & 1 & i & - 1 & - i
e
As argued in the previous parts, if a term number is a multiple of 4 it has a value of 1. The first term number, 396, is a multiple of 4, which means it has a value of 1. With this information, we can also determine the three remaining expressions.
r|ccccc
n & 396^(th) & 397^(th)& 398^(th)& 399^(th)
i^n & 1 & i & - 1 & - i
