Question

use De Moivres theorem to solve the equation
Z^2=-1+sq.root of 3i

Answer 1

do you know what the theorem states?

Answer 2

it is not given to me in the question

Answer 3

But you must have been taught it - otherwise why would you be set such a question?

Answer 4

were were given the following definition

this theorem states that for any complex number (and, in particular, for any real number) x and integer n it holds that
(cos x + i sin x)^n =cos (nx) + i sin (nx)

Answer 5

ok, so this can be stated as follows - if:\[z=r(\cos(\theta)+i\sin(\theta))\]then:\[z^n=r^n(\cos(n\theta)+i\sin(n\theta))\]now, because cos and sin are periodic (i.e repeat every \(2\pi\) radians), we can also say that:\[z=r(\cos(\theta+2\pi k)+i\sin(\theta+2\pi k))\]where \(k=0,1,2,...\).

Does this look familiar at all?

Answer 6

i recognize the first 2 equations as we were only given a brief intro into this modulus. we have to come up with a reasonable solution to compare for tomorrow

Answer 7

ok - but do you follow the reasoning above?

Answer 8

especially about the periodic behaviour of cos and sin.

Answer 9

yes i do

Answer 10

i.e.\[\cos(x)=\cos(x+2\pi)=\cos(x+2\pi*2)=\cos(x+2\pi*3)=...\]\[\sin(x)=\sin(x+2\pi)=\sin(x+2\pi*2)=\sin(x+2\pi*3)=...\]

Answer 11

ok good.

Answer 12

so we can use the theorem to find the \(n\) roots of a complex number as follows...

Answer 13

let:\[z^n=r(\cos(\theta+2\pi k)+i\sin(\theta+2\pi k))\qquad\text{where }k=0,1,2,...\]then we have:\[z=r^{\frac{1}{n}}(\cos(\frac{\theta+2\pi k}{n})+i\sin(\frac{\theta+2\pi k}{n}))\qquad\text{where }k=0,1,2,...,n-1\]this gives the \(n\) roots.

Make sense so far?

Answer 14

yes

Answer 15

great! now lets apply this to your specific question...

Answer 16

you are given:\[z^2=-1+i\sqrt{3}\]the first step is to convert the right-hand-side into the form \(\cos(\theta)+i\sin(\theta)\).

do you know how to do that?

Answer 17

sorry I meant into the form \(r(\cos(\theta)+i\sin(\theta))\)

Answer 18

ok so is it -1(cos(180/-1) + sin(180/rt.3)?

Answer 19

not quite

Answer 20

remember that if you are given a complex number in form \(a+ib\), then we can say that:\[r=\sqrt{a^2+b^2}\]

Answer 21

ok

Answer 22

and:\[\tan(\theta)=\frac{b}{a}\]

Answer 23

so for \(-1+i\sqrt{3}\) we get:\[r=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{4}=2\]and:\[\tan(\theta)=-\frac{\sqrt{3}}{1}=-\sqrt{3}\]therefore:\[\theta=\frac{2\pi}{3}\]

Answer 24

therefore:\[-1+i\sqrt{3}=2(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))\]

Answer 25

make sense?

Answer 26

yes mate thanks a lot

Answer 27

ok, so now we add the periodicity to get:\[-1+i\sqrt{3}=2(\cos(\frac{2\pi}{3}+2\pi k)+i\sin(\frac{2\pi}{3}+2\pi k))\]thus we have:\[z^2=2(\cos(\frac{2\pi}{3}+2\pi k)+i\sin(\frac{2\pi}{3}+2\pi k))\qquad\text{where }k=0,1\]

Answer 28

now we just take square roots of both sides to get:\[z=\sqrt{2}(\cos(\frac{\frac{2\pi}{3}+2\pi k}{2})+i\sin(\frac{\frac{2\pi}{3}+2\pi k}{2}))\qquad\text{where }k=0,1\]

Answer 29

if we call the two roots \(z_0\) and \(z_1\), then we see that:\[z_0=\sqrt{2}(\cos(\frac{\pi}{3}))+i\sin(\frac{\pi}{3}))=\sqrt{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})=\frac{\sqrt{2}}{2}(1+i\sqrt{3})\]

Answer 30

hopefully you can calculate what \(z_1\) is?

Answer 31

ok mate the above is as far as i believe we have to take it for tonight as we are going to be shown the rest tomorrow along with a run through of the above. thanks again mate

Answer 32

yw