Question

16) Find the value of
i⁶+i⁷+i⁸+i⁹/i²+i³

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Answer 1

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Answer 2

Step-by-step explanation:

I don’t quite understand what you mean by ‘all over 1 - i’, but I’ll assume you want to get a result for your problem.

Every power of i corresponds to a distinct number, as you are still essentially multiplying the square root of -1.

i^0 is equal to 1

i^1 is equal to i

i^2 is equal to -1 (the square roots cancel each other)

i^3 is equal to -i (i^3 = i^2 * i = -1 * i)

i^4 is equal to 1

i^5 is equal to i

And so on and so forth. You might notice that after i^3 the numbers start repeating themselves. The first 4 numbers on the list will repeat themselves infinitely as a cycle. For example, i^67 is equal to -i (i^3). To calculate what i^n might be you just have to divide n by 4, and the leftover will be your result. In my previous example, if you divide 67 by 4, you will obtain 16, with a leftover of 3. Thus, i^3.

The answer:

According to all of this you can simplify the problem to:

1 - 1 - i + 1 + i