simplify the power of i: i^-55

Question

sgayton27 · Answer

$$i ^{-55}$$

anonymous · Answer

OGH not these again @ranga

anonymous · Answer

Reaper help i have one more question for you help this one first

anonymous · Answer

mssg me a link

amistre64 · Answer

i is a circular construction, it repeats every 4 times

what is 55/4 ?

amistre64 · Answer

it might be better asked what the remainder is when -55 is divided by 4, but those are semantics

sgayton27 · Answer

3

sgayton27 · Answer

is the remainder

ranga · Answer

i = i
i^2 = -1  
i^3 = -i
i^4 = +1
i^5 = i          (same as first line.  starts repeating)

remainder of 55/4 is 3.  So it will be the third line: -i

i^(-55) = 1 / i^(+55) = 1 / -i = -i / (-i * -i) = -i / +1 = -i

sgayton27 · Answer

i^-55 is the same as $$\frac{ 1 }{ i ^{55} }$$

ranga · Answer

Yes.  See the above answer.

amistre64 · Answer

i^(-3) not= i^(3)

sgayton27 · Answer

well with the remainder being 3 we get $$\frac{ 1 }{ i ^{3} }$$
which is the same as
$$\frac{ 1 }{ 1^{2}*i }$$

which equals
$$\frac{ 1 }{ (-1)*i }$$

so the answer is $$\frac{ 1 }{ -i }$$

amistre64 · Answer

its the same as i^1
    v
  -3  -2  -1
0  1   2   3
4  5   6   7

amistre64 · Answer

the issue with negative exponents is that they are just shy of being expressed as postive exponents.

-3 + 4 = 1 converts it to a usual i^+

ranga · Answer

Oh, I didn't realize the typo in my last line:
i^(-55) = 1 / i^(+55) = 1 / -i = i / (-i * i) = i / +1 = i

ranga · Answer

@sgayton27   Yes, 1 / -i   except you don't leave an i in the denominator.  You rationalize the denominator (meaning you make the denominator a rational number) by multiplying top and bottom by i:$$\frac{ 1 }{ -i } = \frac{ 1 * i }{ -i * i } = \frac{ i }{ -i^2 } = \frac{ i }{ -(-1) }  = \frac{ i }{ +1 } = i$$