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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: fredrick on October 01, 2008, 05:49:56 pm

Title: complex
Post by: fredrick on October 01, 2008, 05:49:56 pm: find all solutions to the equation $z^6=729,z element C$ .

i found the solutions if polar form but the answer and working out is in cartesian form which is messy, can we have the solutions in any form?
Title: Re: complex
Post by: ell on October 01, 2008, 05:57:21 pm: Depends what the question specifies I guess. If it just asks to express your answers in Cartesian I don't see why you wouldn't be able to use De Moivre's theorem then just convert to Cartesian.
Title: Re: complex
Post by: Mao on October 01, 2008, 06:10:37 pm: why would it be messy? they're pi/3 away from each other

$z=\left\{ -3, -\frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}+\frac{\sqrt{3}}{2}i,3,\frac{1}{2}- \frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right\}$
Title: Re: complex
Post by: Collin Li on October 01, 2008, 06:11:19 pm: Yeah - just convert your polar form solutions to cartesian form. Binomial expansion to degree 6 is a waste of time.
Title: Re: complex
Post by: fredrick on October 01, 2008, 06:21:00 pm: so i must have the solutions in cartesian form, but in this question it didnt specify a form.
Title: Re: complex
Post by: shinny on October 01, 2008, 06:23:47 pm: Quote from: fredrick on October 01, 2008, 06:21:00 pm
so i must have the solutions in cartesian form, but in this question it didnt specify a form.

Huh...this is going in circles @.@ If it doesn't specify a particular form, then either's fine...
Title: Re: complex
Post by: Pandemonium on October 01, 2008, 06:44:17 pm: apparently you should put it back into the form that it was first asked for anyway.
but since it's an exam just do that for the ones that matter.