De Moivre's theorem (1 Viewer)

HyperComplexxx · Mar 31, 2011

Find the complex number z which satisfies the equation $z^5 = \overline{z}$
Hints would do

Drongoski · Mar 31, 2011

HyperComplexxx said:
Find the complex number z which satisfies the equation $z^5 = \overline{z}$
Hints would do

Let $z = re^{i \theta}$ or $z = rcis \theta$

HyperComplexxx · Mar 31, 2011

im stumped

Drongoski · Mar 31, 2011

Then

$\bar z = re^{-i \theta} = rcis(- \theta)$

$\therefore (re^{i \theta} )^5 = re^{-i \theta}$

$\therefore r^5 e^{5 \theta i} = r e^{-i \theta}$

$\therefore r^4 e^{6 \theta i} = 1 = e^{0i}$

$\therefore r = 1$ and $6\theta = 0 \pm 2k \pi \text { for k = 0, 1, 2, 3 . . . }$

$\therefore \theta = \pm \frac {k \pi}{3}$

$i.e. z = cis \frac { \pm k \pi}{3} \text { for k = 0, 1, 2, 3 . . . . }$

HyperComplexxx · Mar 31, 2011

$z = 1, \pm \frac{1}{2} (1+i\sqrt{3} ), \pm \frac{1}{2} (1-i\sqrt{3} )$ 0

is that correct?

Drongoski · Mar 31, 2011

I'm too groggy to check; yours look like mine for 1st few cases. What's the "official" answer? You can also check your answers against the original eqn.

blackops23 · Mar 31, 2011

Drongoski said:
Let $z = re^{i \theta}$ or $z = rcis \theta$

WHAT THE HECK IS E^I(THETA)???????

Anways heres wat I did:

Let w = conjugate of z, sorry noob at latex
w=rcis(-A)
let z=rcis(A)
z^5 = r^5(cis5A)

z^5 = w
r^5(cis5A) = rcis(-A)
r^4(cis5A) = cis(-A)
divide by cis(-A) ---> NOT TO SURE IF THIS STEP IS PERMISSIBLE, are you allowed to do this step, or will you be deleting potential solutions?
Assuming I am right...

r^4(cis6A) = 1 = cis(2kpi)

Now i suppose.... r^4 = 1, so r = 1
and 6A = 2kpi????
So A = (kpi)/3???

so z = cis((kpi)/3) k integral

RANK 1 · Mar 31, 2011

e^i(theta) is just cis theta

De Moivre's theorem (1 Viewer)

HyperComplexxx

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blackops23

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