Which of the following is a negative integer if i= #sqrt(-1)#? A) #i^24# B) #i^33# C) #i^46# D) #i^55# E) #i^72#

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Answer 1 · 2018-03-18T18:48:26

#i^46#

#i^1 = i#

#i^2 = sqrt(-1) * sqrt(-1) = -1#

#i^3 = -1 * i = -i#

#i^4 = (i^2)^2 = (-1)^2 = 1#

the powers of #i# are #i, -1, -i, 1#, continuing in a cyclical sequence every #4#th power.

in this set, the only negative integer is #-1#.

for the power of #i# to be a negative integer, the number that #i# is raised to must be #2# more than a multiple of #4#.

#44/4 = 11#

#46 = 44+2#

#i^46 = i^2 = -1#

Answer 2 · 2018-03-21T11:01:03

C) #i^46#

Note that:

#i^0 = 1#

#i^1 = i#

#i^2 = -1#

#i^3 = -i#

#i^4 = 1#

So the increasing powers of #i# will follow a pattern conforming to:

#i^(4k) = 1#

#i^(4k+1) = i#

#i^(4k+2) = -1#

#i^(4k+3) = -i#

for any integer #k#

The only one of these values which is negative is #i^(4k+2) = -1#

Hence the correct answer is C) since #46 = 4*11 + 2#