Question

How do I solve the following question?

Find the indicated roots and graph them in the complex plane.
The cube roots of `1 + i`

Answer 1

See explanation...

Explanation:

de Moivre's formula tells us:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

Note that:

`1 + i = sqrt(2)(cos (pi/4) + i sin (pi/4))`

Hence we find that the cube roots of `1 + i` can be expressed as:

`2^(1/6)(cos (pi/12) + i sin (pi/12)) = 2^(1/6)(1/4(sqrt(6)+sqrt(2))+1/4(sqrt(6)-sqrt(2))i)`

`2^(1/6)(cos ((3pi)/4) + i sin ((3pi)/4)) = 2^(-1/3)(1 - i)`

`2^(1/6)(cos ((17pi)/12)+i sin((17pi)/12)) = 2^(1/6)(-1/4(sqrt(6)-sqrt(2))-1/4(sqrt(6)+sqrt(2))i)`

These roots form the vertices of an equilateral triangle lying on a circle of radius `2^(1/6)` in the complex plane:

graph{(x^2+y^2-2^(1/3))((x-2^(1/6)cos(pi/12))^2+(y-2^(1/6)sin(pi/12))^2-0.004)((x-2^(1/6)cos(9pi/12))^2+(y-2^(1/6)sin(9pi/12))^2-0.004)((x-2^(1/6)cos(17pi/12))^2+(y-2^(1/6)sin(17pi/12))^2-0.004) = 0 [-2.5, 2.5, -1.25, 1.25]}