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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

Title: Urgnt
Post 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
???
Title: Re: Urgnt
Post by: 98.40_for_sure on October 28, 2010, 04:25:13 pm
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]
Title: Re: Urgnt
Post by: SixWinged on October 28, 2010, 04:27:41 pm
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.
Title: Re: Urgnt
Post by: lisafaustina on October 28, 2010, 04:28:58 pm
Ok so you use demoivres and just pick the solutions that are in the domain (-pi,+pi)?
Title: Re: Urgnt
Post by: SixWinged on October 28, 2010, 04:48:00 pm
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.
Title: Re: Urgnt
Post by: plato on October 29, 2010, 12:04:38 am
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.
Title: Re: Urgnt
Post by: chansthename on October 29, 2010, 07:07:30 pm
   


You mean