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Yup. Also, for n > 2 you'd convert to mod-arg form.Probably, but for n > 2 - you'd assume arctan(b/a) was workable.
yerrrrr gud pointWell, just because it appeared in a trial it doesn't necessarily mean that it's in the HSC course... I'll look through past HSC papers from 1990 and see if I find such questions.
I know of course solving z^n = +/- 1 is certainly in the course though. Just not sure about z^n = a+ib.
yep came up in my 4u assessment task todayHi,
I was wondering, are solving equations of the form zn=a+ib (where a,b are real and =/= 0) in the course?
Thanks
what was the exact Q?yep came up in my 4u assessment task today
As I said, just because they put it in an assessment task, it doesn't necessarily mean it is in the course.yep came up in my 4u assessment task today
they do, in the polynomials topic (i think)As I said, just because they put it in an assessment task, it doesn't necessarily mean it is in the course.
And yes I did go through HSC past papers from 1990 and found one (1994 or 1995 IIRC). But it had a lead in part.
So it probably is in the course. But funnily enough I don't think Cambridge goes through it, and I thought that they'd at least cover syllabus material...
I must've missed it then... will have another look tomorrow.they do, in the polynomials topic (i think)
Yeah but some stuff are not in the course. Like for instance partial fractions with repeated factors.anyway, its just an application of de moivre's; i would think that it's an implied part of the syllabus
Even though it is not explicitly stated in the syllabus, it is still assessible in the HSC because you have all the appropriate tools to solve it, without needing anything new or learning about it explicitly. This comes under the "skills" section of the course where you are expected to apply your knowledge of a topic to unfamiliar questions. Generally these types of questions will have a lead in from different parts.Yeah but some stuff are not in the course. Like for instance partial fractions with repeated factors.
UmmShouldn't it just be 2?
|LHS| = 42 + 42 = 16 + 16 = 32
Therefore |z| = fifth root of 32 = 2
i cant remember wat the question was exactly but it was like solve z^6 = a +ib ... a and b were integers and i cant remember them...there were much more harder questions i had to worry about =(what was the exact Q?