Why doesnt this work?

Question

I know if you want to take the cubed root of something you can write it as 8^(1/3) instead of 3radical 8 and samething for the fourth root your can write it as 16^(1/4) instead of 4radical 8 but how come when I want to put (pi*i)radical-1 its not the same as writing -1^(pi*i). In other words how can i write (pi*i)radical -1 in exponential form

I put my calculator in a mode to handle imaginary number and it still doesnt work

Answer 1

because of euler's identity we have
-1 = e^(π*i)

So
(π*i)√(-1) = {e^(π*i)}^(1/(π*i)) = e^1

you can write (π*i)√(-1) as (-1)^(1/(π*i))
and
(-1)^(1/(π*i)) = e = 2.71828182846....

Answer 2

For a positive number x and a positive integer number p, p(radical)x is the positive number y so that y^p = x, i.e. y multiplied by itself p times is equal to x. As a convention, y is written as x^(1/p).

However, for other real numbers p, y^p means exp(p log y), where log is the natural logarithm ("ln" in calculators) and exp is the exponential function ("e^x" in calculators).

If you allow x, y or p to be complex numbers, then there is no unique y that satisfies exp(p log y) = x, so p(radical)x may not be defined, since it can have diferent values. Also, x^(1/p) can have different values, so the equation p(radical)x=x^(1/p) may not be true depending on how you define each side of the equation.

Answer 3

pi*i is another way of saying ln(-1) (if that helps). The reason that you can do pi*i radical (-1) is b/c radical (-1)=i. So you would have pi*i^2 or -pi. If you put -1^(pi*i), the problem would be equal to (1/e)(cospi+isinpi) or something like that. They are 2 different problems. The way you would put i*pi radical -1 in exponential form is (-1)^1/2 * i*pi. or if you are looking for the ipi root of -1, then you would write it like log (-1)x=ln (-1).

Answer 4

You would need to write the exponential form as
(-1) ^ ( 1 / (pi*i))

Answer 5

the square root of -1 is not the same as just -1 by itself, whereas all the other examples it meant the same thing in the expression (maybe)

Answer 6

the domain does not include negatives.
For example, you can not do square root of (-4)
If the base is negative, we have restrictions on the exponent, it canot be (pi*i)
what would (-1)^(pi) be?? we cannot even determine its sign.

Answer 7

pi*i*sqrt(-1) = pi*i^2 = e^ln(pi*i^2) =e^ln(-pi)
but -pi = pi (Principal determination as ln of complex numbers has infinite
values
=e^(ln pi+i*pi)

Answer 8

If you are trying this on your calculator, it isnt working because your calculator isnt programmed to do any negative powers because it is impossible.

Answer 9

If you're trying to take the "πi"th root of -1, then that would be
(-1)^(1 / πi), not (-1)^(πi).

Answer 10

when you have a negative square root u have to use imaginery numbers i