ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: lisafaustina on October 28, 2010, 04:23:36 pm
-
Ok so you know how they ask you in complex numbers to findthe solutions to z^4=9 or something like that, which domain to you use? -pi to +pi or 0 to 2pi
???
-
You don't use a domain.
You convert it to polar form then apply de moivre's theorem.
Just remember to add +2kpi before taking it to the ^(1/4)
You will just get 4 solutions before they repeat.
Simplify the angles to (-pi,pi]
-
Either method is fine for finding the answers, in fact the method you've probably been taught involves adding 2pi(n) where n is an element of J. However the proper way to express a complex number is with an Argument of (-pi, +pi]. So once you're done working the correct answers you can add and subtract 2pi until your answers are within the expected domain.
-
Ok so you use demoivres and just pick the solutions that are in the domain (-pi,+pi)?
-
Yeah, every other solution is just a repeat of one of those solutions anyway since it will differ by 2pi, meaning it's not a unique solution.
-
Ok so you know how they ask you in complex numbers to findthe solutions to z^4=9 or something like that, which domain to you use? -pi to +pi or 0 to 2pi
???
For an equation like
, you could take the square root of both sides to get

Then
or 
The four solutions are equally spaced at intervals of
around a circle of radius
on the complex plane
The complex roots are conjugates.
-

You mean