Edexcel FP2 complex numbers HELP!

Hi I need help on the last part of this question:

z = − 8 +(8√3)i
(a) Find the modulus of z and the argument of z.

angle = 2pi/3 using tan, and r = 16


Using de Moivre’s theorem,

(b) find z^3

z^3 = 4096 ( cos 2pi + i sin 2pi ) using the formula for de moivres.


(c) find the values of w such that w^4 = z, giving your answers in the form a + ib, where a b, ∈.

I have found w to be z^0.25 == 2 ( cos pi/6 + i sin pi/6 ) meaning
w = √3 + i.

but answers also state: OR −1 + √3i OR −1 − √3i OR 1 + √3i.

Can someone please tell me how the last 3 are obtained? the link to markscheme is here: it is question 4.

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-ms/6668_01_msc_20100714.pdf

Thanks in advance!

hello. When we're finding solutions for problems like these, we need to recognise that the argument (the multiple of pi inside cos and sin) is not unique. If you look at a sin or cosine graph, you'll see that values of $\theta$ are repeated every 2pi. In order to account for this, we need to say:

$w = [16(cos(\frac {2\pi} {3} + 2k\pi) + isin(\frac {2\pi} {3} + 2k\pi))]^{\frac {1} {4}}$

where $2k\pi$ is simply a multiple of $2\pi$ (k is always a positive or negative integer).

Evaluating this using de moivre's theorem:

$w = 2(cos(\frac {\pi} {6} + \frac {k\pi} {2}) + isin(\frac {\pi} {6} + \frac {k\pi} {2}))$

now, we need to substitute certain values of k, and this will give us several arguments. We're only looking for arguments between $-\pi$ and $\pi$. You'll want to work your way up from 0, then to 1 and -1, then to 2 and -2, then to 3 and -3, etc... So let's say:

$\theta = (\frac {\pi} {6} + \frac {k\pi} {2})$.

When k=0, $\theta = \frac {\pi} {6}$

when k=1, $\theta = \frac {2\pi} {3}$

when k=-1, $\theta = \frac {-\pi} {3}$

when k=2, $\theta = \frac {7\pi} {6}$. This is larger than ${\pi}$ so we discard it.

When k=-2, $\theta = \frac {-5\pi} {6}$

every value of k above 1 and below -2 will now give an argument outside the range of $-\pi$ to $\pi$.

Therefore, the suitable arguments are $\theta = \frac {\pi} {6}, \frac {2\pi} {3}, \frac {-\pi} {3}, \frac {-5\pi} {6}$

by substituting all of these into:

$w = 2(cos(\theta) + isin(\theta))$

we get: $w=\sqrt 3 + i$
$w=-1 + i\sqrt 3$
$w=1-i\sqrt 3$
$w=-\sqrt 3 - i$

this is how i do these sorts of questions.

$z=16e^\frac{2\pi i}{3}$ and we want $w: W^4=z$

if we let $w=re^{i\theta}$ then $w^4=z \implies w^4 = r^4e^{4\theta i}=16e^\frac{2\pi i}{3}$. Now comparing absolute value and arguments, we get $r=4$ and $4\theta = \frac{2\pi}{3}+2k\pi$ for some integer k. We do this because the argument of a complex number isn't unique. In the end, $\theta = \dfrac{\pi}{6}+\dfrac{k\pi}{2}$ for k=0,1,2,3 (of course if k=4 then we get the same number as when k=0, etc.) are the 4 different arguments that w can take, and hence give the 4 complex numbers you're looking for.