complex.... (1 Viewer)

[*fkr]

hey,
im just working through finding the roots of complex num's and am having problems getting all the right answers for 4 roots and more. could someone please show me the process they use for:
find the five 5th roots of: - √3 / 2 + 1 / 2 (patel 4e-4c)
cheers

 

[pLuvia]

- √3 / 2 + 1 / 2

is it suppose to be

- √3/2 + i(1/2)?

 

[*fkr]

nope......

 

[100percent]

hey,
im just working through finding the roots of complex num's and am having problems getting all the right answers for 4 roots and more. could someone please show me the process they use for:
find the five 5th roots of: - √3 / 2 + 1 / 2 (patel 4e-4c)
cheers

- √3 / 2 + 1 / 2 <--that isn't complex number

 

[Slidey]

- √3 / 2 + 1 / 2
in Rcis@ form it is:
(sqrt3 - 1)/2 * cis(pi)

Ignoring the modulus for now:

z^5=Rcis(pi)
z=rcis([pi+2kpi]/5)
k=0 produces z₁=rcis(pi/5)
k=1, z₂=rcis(3pi/5)
k=2, z₃=-r
k=3, z₄=rcis(-3pi/5)
k=4, z₅=rcis(-pi/5)
where r=((sqrt3 - 1)/2)^(1/5)

Now multiply all of those roots by the modulus.

 

[Slidey]

- √3 / 2 + 1 / 2 <--that isn't complex number

Yes it is. The set of real numbers is a member of the set of complex numbers.

 

[100percent]

Yes it is. The set of real numbers is a member of the set of complex numbers.

ok, my bad..explains why i almost failed 4unit

 

[Riviet]

- √3 / 2 + 1 / 2 <--that isn't complex number

In addition to slide rule's defintion: - √3 / 2 + 1 / 2 = - √3 / 2 + 1 / 2 + 0i, which is a complex number but where there is no imaginary part to the number.

 

[mitsui]

woot.. cant understand wat every1 is on about. xD

 

[Riviet]

woot.. cant understand wat every1 is on about. xD

LOL! Have you started complex numbers yet? Nothing to be "wooting" about,

 

[noah]

- √3 / 2 + 1 / 2
in Rcis@ form it is:
(sqrt3 - 1)/2 * cis(pi)

Ignoring the modulus for now:

z^5=Rcis(pi)
z=rcis([pi+2kpi]/5)
k=0 produces z₁=rcis(pi/5)
k=1, z₂=rcis(3pi/5)
k=2, z₃=-r
k=3, z₄=rcis(-3pi/5)
k=4, z₅=rcis(-pi/5)
where r=((sqrt3 - 1)/2)^(1/5)

Now multiply all of those roots by the modulus.

do you not also have to take the 5th root of the modulus?

EDIT: sorry ignore me. i didnt read the post carefully enough

 

[Slidey]

It was a poorly constructed post. You won't get an ugly question like that in the HSC.

They can certainly ask for 5th roots but I imagine you'd get a nice modulus and a nice argument.

 

[Raginsheep]

Its always good to practice i guess. Who knows, you might get thrown one of these as part (a) (i) of question 8...

 

[*fkr]

i asked my teacher today and he said he photocopied the unedited version which has the wrong question for the answer, kadlil was right... but thanks anyway

 

[pLuvia]

hehe, still Slide could work it out haha, and I didn't take into account that the complex number was every number but now I know