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

2 Answers

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}

Jun 28, 2018

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.