Question

a solution for z^3 +2 + 2 sqrt3i = 0, by using the n^th root formula for complex numbers is?

Answer 1

go to this link : http://en.wikipedia.org/wiki/Square_root

Answer 2

Thanks, but its not helping much

Answer 3

express the complex number in the form:
\[e ^{i \theta}\]

Answer 4

using eulers formula we have
\[R e^{i\theta} =R( \cos\theta + isin\theta) \]
hence
\[(R e^{i\theta})^n = R^n e^{i n\theta} = R^n(cosn\theta + isinn\theta)\]

in your case n = 3
so express z^3 as

\[R^3(cos3\theta + isin3\theta) = -2 -2\sqrt{3}i \]

now solve for R and theta

Answer 5

taking the modulus of both sides:

\[R^3 = \sqrt{2^2 + (2\sqrt{3})^2}\]

Answer 6

and finding the argument:

\[tan3\theta = \frac{-2\sqrt{3}}{-2} = \sqrt{3}\]

Answer 7

do you follow?

Answer 8

thanks, but I'm completely clueless... think I need some background, any suggestions? I have possible answers, see attached

Answer 9

what background knowledge do you have on this already?

Answer 10

complex numbers i mean

Answer 11

not much hey, basically what a complex number is and some formulas to use:
|a+bi| = \[\sqrt{a^2+b^2}\]

a=r cos \[\theta\]

b = r sin \[\theta\]

a + bi =r\[\left( \cos \theta + i \sin \theta \right)\]

Answer 12

ok, have you seen the geometrical effect of complex multiplication?

try this:

let\[z_1 = 1+i \text{ , }z_2 = 2+ \sqrt{3}i\]

find the argument and modulus of each z and then find the argument and modulus of

\[z_1z_2 = (1+i)(2+\sqrt{3}i)\]

see if you can work out the relationship

Answer 13

if you have come across compound angle formulae, try doing this in general with:
\[z_1 = \cos\theta + i \sin\theta \text{ , } z_2 = \cos\phi + i \sin\phi\]

Answer 14

remember all the while what you are doing geometrically:

Answer 15

ok, give me some time to work on this, will reply a bit later...

Answer 16

sure :) i love this area of maths, where complex and trig overlap

Answer 17

maybe I'll get there one day:-)

Answer 18

Hi, still couldn't figure this one out so will tackle it some time later again. Thanks anyway.