If you consider the complex z-plane the answer is immediately clear : for the 1/3 -power you have to divide the corresponding angle by 3: π/2 * 1/3 = π/6 :hence the answer is cos π/6 +i*sin π/6 =1/2*(√3 + i). In your video you give the solution of z^3 = i, but the third root is uniquely defined, as the solution given here.
i=e^(i*Pi/2+k*2*Pi) k=0; +/-1; +/-2; ... -> 3rd-root(i)=e^(i*Pi/6 + k*2*Pi/3); k=0 -> sqrt(3)/2 + i/2; k=1 -> -sqrt(3)/2 +i/2; k=2 -> -i; k=3 is the same result for k=0; k=4 results the same as k=1; Knowledge to solve: e^(ix) = cos(x) + i*sin(x) and sin(30°)=1/2; sin(60°)=sqrt(3)/2=cos(30°) and x = Pi/6 is equavalent to 30°
If you consider the complex z-plane the answer is immediately clear : for the 1/3 -power you have to divide the corresponding
angle by 3: π/2 * 1/3 = π/6 :hence the answer is cos π/6 +i*sin π/6 =1/2*(√3 + i). In your video you give the solution of z^3 = i,
but the third root is uniquely defined, as the solution given here.
{i+i+ii ➖ i ➖ i}=i^3 (x➖ 3ix+3i).
i=e^(i*Pi/2+k*2*Pi) k=0; +/-1; +/-2; ... -> 3rd-root(i)=e^(i*Pi/6 + k*2*Pi/3); k=0 -> sqrt(3)/2 + i/2; k=1 -> -sqrt(3)/2 +i/2; k=2 -> -i; k=3 is the same result for k=0; k=4 results the same as k=1; Knowledge to solve: e^(ix) = cos(x) + i*sin(x) and sin(30°)=1/2; sin(60°)=sqrt(3)/2=cos(30°) and x = Pi/6 is equavalent to 30°
I'd rather use the De Moivre's formula.
i³=-root-1 right
³√i = i^(1/3) = (e^((π/2+2kπ)*i))^(1/3) = e^((1+4k)πi/6) k ∈ Z
for k=0: ³√i = e^(πi/6) = cos(π/6) + i*sin(π/6) = √3/2 + i*1/2 = (√3+i)/2
for k=1: ³√i = e^(5πi/6) = cos(5π/6) + i*sin(5π/6) = -√3/2 + i*1/2 = (-√3+i)/2
for k=2: ³√i = e^(9πi/6) = cos(3π/2) + i*sin(3π/2) = 0 + i*(-1) = -i