Is the square root of "i", i.e.(-1)^(1/4), a number that exist?
回答 (7)
2006-01-24 1:34 pm
✔ 最佳答案
i = cos(pi/2) + isin(pi/2)sqrt(i) = cos(pi/4) + i sin(pi/4) = sqrt(2)/2 + i sqrt(2)/2
if z = r(cos(t) + i sin(t), to take the square root, you take the square root of r (in your case 1) and divide t by 2.
2006-01-25 1:22 pm
sqrt(i) = .707 + .707i
2006-01-24 10:22 pm
I agree with rt11guru
2006-01-24 10:12 pm
"Imaginary" numbers are a misnomer and the term causes much confusion. The complex numbers exist as an "extension" of the reals.
You need to define what a number is. Does the number "1" exist? There is after all no existence of a "1" on the earth - so it too is, in a sense, imaginary (only a construct in our heads)
What I think your question might be is "Is there a number in the complex numbers which is equal to the square root of i"? The answer then yes e.g. solve
(a+bi)^2=i and you'll find one possible solution is 2^(-1/2) + 2^(-1/2)i, just like the first answer from rt11guru.
You need to define what a number is. Does the number "1" exist? There is after all no existence of a "1" on the earth - so it too is, in a sense, imaginary (only a construct in our heads)
What I think your question might be is "Is there a number in the complex numbers which is equal to the square root of i"? The answer then yes e.g. solve
(a+bi)^2=i and you'll find one possible solution is 2^(-1/2) + 2^(-1/2)i, just like the first answer from rt11guru.
2006-01-24 9:45 pm
Imaginary numbers are those which doe not exist. So this one can also be treated as an imaginary number
參考: Abraham's extensive mathematics. By pro Sunny Abrahanm. Prof in IIT
2006-01-24 9:43 pm
i = root (-1) itself is an imaginary number...it does not exist. so root(i) definitely doesnot exist.
2006-01-24 9:36 pm
If you want numbers that are not in the Argand Diagram, you might consider quaternions and Cayley numbers.
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