Question

help.
cube root of -2i in r (costheta +isintheta) form?

Answer 1

let z = -2i
x = 0 , y = -2
theta = -2/0 , theta = 270

|r| = sqrt( 0 + 4) = 2

z = |r| (costheta + sintheta) = 2 ( cos270 + i(sin270))

You can easily verify it, by calculating the value of sin , cos, multiplying times the 2 out.

Answer 2

thank you!

Answer 3

how would I write it with cuberoot(2) as my r value?

Answer 4

you should have 3 cube roots

Answer 5

nevermind, I think I figured it out, since cbrt(2) = 2^(1/3), all I do is divide 270 by 3 and get cbrt(2)(cos(90)+isin(90))

Answer 6

or not, it appears it was 210 :(

Answer 7

\(\large -2i = 2 e^{i \frac{3 \pi}{2}}\) but also \(\large -2i = 2 e^{i (\frac{3 \pi}{2}+2 \pi)}\), \(\large -2i = 2 e^{i (\frac{3 \pi}{2}+4 \pi)}\) , ......

so \(\large -2i = 2e^{i\frac{3 \pi}{2}}, 2e^{i\frac{7 \pi}{2}} , 2e^{i\frac{11 \pi}{2}} , 2e^{i\frac{15 \pi}{2}},....\)

so \(\sqrt[3]{-2i } = \sqrt[3]{2}e^{i\frac{ \pi}{2}}, \sqrt[3]{2}e^{i\frac{7 \pi}{6}} , \sqrt[3]{2}e^{i\frac{11 \pi}{6}} , \sqrt[3]{2}e^{i\frac{15 \pi}{6}},....\)

so the fourth root is a repeat of the first as \( \frac{15 \pi}{6} = \frac{\pi }{2} + 2 \pi \); and they repeat ever after....

so in cis form you have \(\sqrt[3]{-2i } = \sqrt[3]{2}\; cis \, 90, \sqrt[3]{2} \; cis\, 210 , \sqrt[3]{2} \; cis \,330 ,\)

Answer 8

but the 90deg one would be the *principal* cube root of you are looking for only one answer....