This is equal to 1. On the Wikipedia page for imaginary numbers, they have a table, but here is a summary for in:
n value of i^n
-- ------
-4, 1
-3, i
-2, -1
-1, -i
0, 1
1, i
2, -1
3, -i
4, 1
Notice there is a repeating pattern.
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Any real number to the power of '0' is '1' :: i to the power of 0 = 1
I believe that to be a correct statement.
complex numbers with no real partif any complex number z can be written a + i bthen pure imaginary numbers have a=0 and b not equal to 0
A pure imaginary number is a complex number that has 0 for its real part, such as 0+7i.
Both imaginary and real numbers are infinite .Answer:Any real number can be turned into an imaginary number by multiplying it by "i" ot "j" (the root of -1). Hence it would appear that the set of all real numbers would equal the set of all imaginary numbers. However 0 (zero) multiplied by anything still equals zero. This would mean that there is at least one number that cannot be converted to an imaginary number.
No. A complex number consists of a real part and a imaginary part. If the real part equals zero, there is only the imaginary left and you could therefor argue that it is an imaginary number (or else it would still be a complex number -with a real part=0)
Yes, the only argument would be the example, i + (-i) = 0. However, many people don't realize that 0 is both a purely real and pure imaginary number since it lies on both axes of the complex plane.