I can't do this fecking question!! I'm probably missing something really obvious but...well here it is

z^3 = i

where i = sqrt (-1)

Solve for all values of 'z'

Can anyone show me how to do it? :(

Edit: Ok, I think I got it but post here anyways just to make sure I am right

Note that i has several representations as a complex exponential: e^(i*(pi+4*n*pi)/2) where n is an integer 0, 1, 2, 3, ...

Now i^(1/3) = (e^(i*(pi+4*n*pi)/2))^(1/3) = e^(i*(pi+4*n*pi)/6). Since this is a periodic function, one only needs to check all the values of (pi+4*n*pi)/6 that are less than 2*pi - then the values just repeat ad infinitum.

e^(i*pi/6) = cos(pi/6) + i*sin(pi/6) = sqrt(3)/2 + (1/2)*i

e^(i*5*pi/6) = cos(5*pi/6) + i*sin(5*pi/6) = -sqrt(3)/2 + (1/2)*i

e^(i*9*pi/6) = cos(3*pi/2) + i*sin(3*pi/2) = 0 + -i

The next values for i^(1/3) is e^(i*13*pi/6), which is exactly the same as e^(i*pi/6), so the values will repeat after this.

Solutions:

sqrt(3)/2 + (1/2)*i
-sqrt(3)/2 + (1/2)*i
-i

z^3=i
z^3=(-1)^(1/2)
z=(-1)^(1/6)
z=((-1)^(1/3))^(1/2)
z=(-1)^(1/2)
z=i


that doesnt work..... i^3=-i
oh well

Note that i has several representations as a complex exponential: e^(i*(pi+4*n*pi)/2) where n is an integer 0, 1, 2, 3, ...

Now i^(1/3) = (e^(i*(pi+4*n*pi)/2))^(1/3) = e^(i*(pi+4*n*pi)/6). Since this is a periodic function, one only needs to check all the values of (pi+4*n*pi)/6 that are less than 2*pi - then the values just repeat ad infinitum.

e^(i*pi/6) = cos(pi/6) + i*sin(pi/6) = sqrt(3)/2 + (1/2)*i

e^(i*5*pi/6) = cos(5*pi/6) + i*sin(5*pi/6) = -sqrt(3)/2 + (1/2)*i

e^(i*9*pi/6) = cos(3*pi/2) + i*sin(3*pi/2) = 0 + -i

The next values for i^(1/3) is e^(i*13*pi/6), which is exactly the same as e^(i*pi/6), so the values will repeat after this.

Solutions:

sqrt(3)/2 + (1/2)*i
-sqrt(3)/2 + (1/2)*i
-i

Wow.
I have no idea what the fuck that is.
+1


I got as far as:

z^3=i
z^3= sqrt(-1)
z=cuberoot(sqrt(-1))

Then I gave up because I have no idea what I'm doing.

When dealing with roots and complex numbers you have to be careful like Shadout Mapes did with using Euler's Formula and polar representations, because its really easy to make a mistake otherwise. Confusion arises easily with the multi-valuedness, and many of the rules of roots that are true in real numbers are not true with complex numbers, like for instance with HeaT's post, mistakes arise if you try to use the rules of real numbers. I think the easiest way to understand and deal with complex numbers is with polar representation, which is worth it learn well.

Sweet, thanks. From memory that's what I got in the end...although I don't have my assignment here so I don't know for sure. But thanks for the help!