What is the answer?

Imaginary numbers:

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2 Answers
Nov 14, 2017

#i^0 = 1#
#i^1 = i#
#i^2 = -1#
#i^3 = -i#
#i^4 = 1#

Explanation:

Generally if you want to get the solution to #i^x# where #x# is any non-negative integer, you need to remember the following:

#i^0 = 1#
#i^1 = i#
#i^2 = -1#
#i^3 = -i#
#i^4 = 1#

For all non-negative integers, if it is an exponent of #i# then you simply have to divide #x# by 4, 3, or 2.

If #x# is divisible by 4 then #i^x = 1#
If #x# is divisible by 3 then #i^x = -i#
If #x# is divisible by 2 then #i^x = -1#
If #x# is not divisible by any of the above numbers then #i^x = i#
If #x# is 0 then #i^x = 1# since anything raised to 0 is equal to 1
Note that for this you must choose the highest number among 4, 3, and 2 that #x# is divisible with.

Similarly,
#i^2 = i * i = -1#, so it also goes that:
#i^3 = i * i * i = i^2 * i = -1 * i = -i#
#i^4 = i * i * i * i = i^2 * i^2 = -1 * -1 = 1#

Nov 14, 2017

#i^0=1#
#i^1=i=sqrt(-1)#
#i^2=-1#
#i^3=-i=-sqrt(-1)#
#i^4=1#

Explanation:

#i=sqrt(-1)#

#i^0=(x^0=1 | x in C)# ... any number (x) power 0 is 1
#i^1=i# ... any number (x) power 1 is the number (x)

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

#i^3=i*i^2=sqrt(-1)*1=sqrt(-1)=i#

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


note: for any #i^a=b#, one can write
#(i^(2n))(i^(a-2n))=b#
when
#2n# is Even
AND
#2n-a=1 or 2n-a=3#

for exmple:
#i^35=?#

We will take #(2n=32,2n-a=3)#

#i^35=i^32*i^3=((i^4)^8)*i^3=(1)^8*i^3=1*i^3=i^3=i#

end note