2i^7
i^42
i^100
I assume you want to simplify each one
the imaginary number i has very unusual properties
watch this cyclic pattern:
i
i^2 = -1
i^3 = i(i^2) = -i
i^4 = i^2 i^2 = (-1)(-1) = +1
i^5 = i i^4 = i(1) = i , and here we go round again
notice every even power of i is either +1 or -1
if the even power is divisible by 4 we have +1
if the even power is NOT divisible by 4, e.g. 6 , we get -1
if dividing the exponent by 4 leaves a remainder of 1 we get +i
if dividing the exponent by 4 leaves a remainder of 3, we get -i
so i^42 ???
42 ÷ 4 is not exact, but 42 is even , so
i^42 = -1
of course 2 i^7 will give us 2(-i) = - 2i
So what do you think about i^100 ?
so on 2^7i list i as -1 seven times equals out -1 or -i then what
i still cant solve i^100
were did you get 2^7i from ???
I thought we did 2 i^7
and I thought I did that for you.
i^100
= (i^4)^25
= (+1)^25
= 1
I strongly suggest your write down the first 4 or 5 terms of the pattern I gave you and I strongly suggest you read my explanation of that pattern.
To find the values of powers of i, we need to understand the pattern of i raised to various powers.
The imaginary unit i is defined as the square root of -1. Therefore, we can establish the following pattern:
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
The pattern then repeats every four powers. So, to solve the given questions:
1. 2i^7:
Since 7 is not divisible by 4, we can divide it by 4 to find out the remainder:
7 ÷ 4 = 1 remainder 3
Now, we can substitute back the remainder into the pattern to find the value:
i^3 = -i
Thus, 2i^7 = 2(-i) = -2i.
2. i^42:
Again, we divide 42 by 4 to find out the remainder:
42 ÷ 4 = 10 remainder 2
Substitute the remainder into the pattern:
i^2 = -1
So, i^42 = -1.
3. i^100:
By dividing 100 by 4, we get:
100 ÷ 4 = 25 remainder 0
When the remainder is 0, it means the power is divisible by 4, and the pattern of i^4 = 1 applies.
Therefore, i^100 = 1.
In summary:
- 2i^7 = -2i
- i^42 = -1
- i^100 = 1