Question

How do you find the cube root of `-125/2(1+sqrt3i)`?

Answer 1

`5e^{-{2pi i}/9}, \ 5e^{-{4pi i}/9}, \ 5e^{-{10pi i}/9}`

Explanation:

`-125/2(1+\sqrt3i)`

`=-125/2-{125\sqrt3}/2i`

`=125e^{-{2pi}/3 i}`

Now, cubic roots of `-125/2(1+\sqrt3i)` are given as

`(125e^{-{2pi}/3 i})^{1/3}`

`=5(e^{i(2k\pi-{2pi}/3)})^{1/3}`

`=5e^{i/3(2k\pi-{2pi}/3)}`

`=5e^{i({2k\pi}/{3}-{2pi}/9)}`

Where, `k=0, 1, 2`

Thus, setting `k=0,1, 2` in above general cubic root, we get three cubic roots of given complex number

`\root[3]{-125/2(1+\sqrt3i)}=5e^{-{2pi i}/9}, \ 5e^{-{4pi i}/9}, \ 5e^{-{10pi i}/9}`

Answer 2

There are three cube roots,

`-5 (cos (-{5pi}/9) + i sin (-{5pi}/9))`

`-5 (cos (pi/9) + i sin (5pi/9))`

`-5 (cos ({7pi}/9) + i sin ({7pi}/9))`

Explanation:

`z = ( -125/2(1+sqrt{3} i))^(1/3)`

`= (-125)^{1/3} (1/2 + sqrt{3}/2 i )^{1/3}`

`= -5 (cos pi/3 + i sin pi/3)^{1/3}`

`= -5 (cos (pi/3+ 2pi k) + i sin (pi/3 + 2 pi k) )^{1/3} quad` integer `k`

We add the `2pi k` because every complex number has three cube roots and we want to find them all.

De Moivre's theorem works when `n` is a fraction too; it's basically Euler's Formula.

`= -5 (cos (pi/9 + {2pi k}/3) + i sin (pi/9 + {2pi k}/3))`

That's three unique values, given by any three consecutive `k.` These aren't constructible angles so there's no nice expression combining integers, addition, subtraction, multiplication, division and square rooting. We'll just write the three possibilities, `k=-1,0,1.`

`z = -5 (cos (-{5pi}/9) + i sin (-{5pi}/9))` or

`z = -5 (cos (pi/9) + i sin (5pi/9))` or

`z = -5 (cos ({7pi}/9) + i sin ({7pi}/9))`

We can lose the minus sign by adding `pi` to the angles, but I won't bother.