Question

How do i find the solutions for x^3+125i=0, using the roots of a complex number theorem?

Answer 1

\[n \sqrt{r}(\cos (\theta +2 pi k)/n + i \sin (\theta + 2 pi k)/n) \]

Answer 2

Y U NO ...? (I can't read it :/)

Answer 3

Math y u no be easy?

Answer 4

\[x^3 = -125i \implies x = \sqrt[3]{-125i} \implies x = \large -5e^{i\frac{\frac{\pi}2 + 2n\pi}3}\]

Answer 5

Lol.

Answer 6

so the standard form is going to be 0-5i?

Answer 7

\[\frac{\pi}6 + \frac23n\pi < 2\pi\]
\[n < \frac{11}4,\quad or\quad n \le 2\]

Answer 8

idk what the standard form is. :/ but yeah cos\theta has to be zero.

Answer 9

I see and since it only goes in the imaginary direction it will be pi/2? from there I plug 3 into n in the equation?

Answer 10

Yes. From there you take a round in the unit circle and find all the values for theta 0<theta<2\pi.

Answer 11

Ok thanks a lot, you truly are a sir.

Answer 12

You're welcome :-)