Complex Roots (1 Viewer)

[ezzy85]

express roots of z⁵ + 32 = 0 in mod/arg form:

Heres what im getting:

2cis@ where @ = pi/5, 3pi/5, 7pi/5, 9pi/5

The part im confused about is the last 2 thetas and getting them into the required range. is 7pi/5 = -3pi/5 and then do you just sub that in to get 2cis(-3pi/5) = 2cos 3pi/5 - 2i sin 3pi/5 or is it:

-2cos 3pi/5 - 2i sin 3pi/5 since both cos and sine are negative in the third quadrant.

I always get confused where the minus goes.
Thanks

 

[Constip8edSkunk]

cos (-3pi/5) = cos (3pi/5) < 0
but
-cos(3pi/5) > 0

 

[ND]

Originally posted by ezzy85
is 7pi/5 = -3pi/5 and then do you just sub that in to get 2cis(-3pi/5) = 2cos 3pi/5 - 2i sin 3pi/5

That's right.

 

[freaking_out]

hey is there anything wrong with using that formula for roots of complex no. in fitzpatrick???

if u don't know, i'll post it up.

 

[ezzy85]

please post it up

 

[Rahul]

umm....why dont u use the circle of unity to solve it?

set ur first root at -2, then rotate each of the other four roots around the circle by 2pi/5. that will give you, (+-) cis pi/5 and (+-) 3pi/5.

if you draw this on the circle you will find that you wont have to worry about the range and repeating the angles.

 

[freaking_out]

yeah, it just a general formula...look at attachment
(its from fitzpatrick)... note the k is just like the parameters u sub in when u do it normally.

 

[Richard Lee]

The formula is:
cos(-theta)=cos(theta)
Don't worry about the value of the theta. That's rules.
eg. cos(-220 degree)=cos(220 degree)
and cos(-1030 degree)= cos(1030 degree)