Question

I*(-1)^11 ...What does this simplify to?
and what are the powers of "i"?
Also, Can you give me two more similar problems to i^23 and explain how to do it, to get an idea.

Answer 1

@BangkokGarrett Can you help me?

Answer 2

Is that an i times -1 raised to the 11th power ???

Answer 3

yes it is

Answer 4

well..(-1)^11 = -1.......-1 ^ any odd number will equal -1
then, i times -1 will just be -i. So the answer is -i .

Answer 5

i = i
i^2 = -1
i^3 = -i
i^4 = 1
Then it repeats the same cycle...
i^5 = i
i^6 = -1
etc etc.

Answer 6

To find a big one like i^23...find the nearest multiple of 4
24 is a multiple of 4, right?
So, we know i^24 = 1....since it goes in cycles of 4 like the pattern I just showed you.
Now just count backwards in the cycle to get i^23
We just need to count back 1 .. so i^23 = -i

Answer 7

oh okaay thankyou!!!

Answer 8

i^102
100 is the nearest multiple of 4
so i^100 = 1
now just count forward
i^101 = i
i^102 = -1

Answer 9

can you give me one more like that please and How do I simplify \[x=-2\pm \sqrt{-6}\]?

Answer 10

\[\sqrt{-6}=\sqrt{-1\times 6}=\sqrt{-1}\sqrt6=6i\]

Answer 11

thankyou can you do me one more favor?