Complex numbers (1 Viewer)

[Teoh]

Troubles with:

Finding the 5th roots of 1...



And also:

z = cos? + isin?, show that zⁿ+ 1/zⁿ= 2cos?

 

[Ragerunner]

Modulus = 1
Argument = 0

1 = 1e^2kipi

r^5 = 1

5@ = 2kpi

@ = 2kpi/5

k = -2,-1,0,1,2

Just substitute the k values into 2kpi/5

e.g. for k = -2

@ = -4pi/5

Then in polar form is e^-4pi/5

And do it for the rest of the values of k.

 

[wogboy]

If you're wondering what e^ix means,

e^ix = cosx + isinx = cisx
(don't worry why this is true, it isn't in the HSC syllabus)

Some people (particularly uni lecturers) prefer writing e^ix instead of cisx.

z = cos + isin, show that z^n+ 1/z^n= 2cos

There's probably a misprint, it should be:
show that z^n+ 1/z^n= 2cos(n)

To show that, use De Moivre's theorem, bearing in mind that 1/z^n = z^-n.

 

[Teoh]

Originally posted by Ragerunner
Modulus = 1
Argument = 0

1 = 1e^2kipi





So that's just
1 = 1cis2kpi



Thanks for the help.

 

[Ragerunner]

Yes it is.

 

[KeypadSDM]

Originally posted by Teoh
1 = 1cis2kpi

Please, please use brackets.