Which ordered pairs are solutions to the equation?
1 answer:
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Problem 5
<h3>Answer: 6i</h3>------------------
Explanation:
We have these four identities
- i^0 = 1
- i^1 = i
- i^2 = -1
- i^3 = -i
Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.
We can think of it like this if you wanted
i^28 = (i^4)^7 = 1^7 = 1
Then the sqrt(-36) becomes 6i
So overall, we end up with the final answer of 6i
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Problem 6
<h3>Answer: -3i</h3>------------------
Explanation:
We'll use the ideas mentioned in problem 5
46/4 = 11 remainder 2
i^46 = i^2 = -1
sqrt(-9) = 3i
The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i
In other words, here is one way you could write out the steps
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Problem 7
<h3>Answer: -1</h3>------------------
Work Shown:
i^10 = i^2 because 10/4 = 2 remainder 2
i^19 = i^3 because 19/4 = 4 remainder 3
i^7 = i^3 because 7/4 = 1 remainder 3
Again, all we care about are the remainders.
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Problem 8
<h3>Answer: -1 + i</h3>------------------
Work Shown:
i^22*i^6 = i^(22+6) = i^28
Earlier in problem 5, we found that i^28 = i^0 = 1
So,