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doing questions another way (1 Viewer)

mojako

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will we still get full marks for this?
Say for example, HSC 1998, Q2 (e) (i)
By solving the equation z^3 + 1 = 0, fnd the three cube roots of -1.

If I do it by solving
z^3 = -1, -1 = cis(pi),
using De Moivre's theorem,
will that be ok?

I think you're meant to solve (z+1)(z^2 - z + 1) = 0
 

mojako

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hmm after seeing part (ii) I think you should use De Moivre's

ok, so say you're meant to use De Moivre's in part (i)
But I solve the quadratic like above,
is it ok?
 

grimreaper

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I'm certain thatd be alright... its the way I've always done questions like that (in reference to using de moivres)
 

CrashOveride

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the suggested solution is to form the factors and solve. but nothing wrong with using DM
 

mojako

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what about things like this:
integral of sin^3x/cos^2x.dx

you get "different" answers using different method...
the normal one is
sec(x) + cos(x)

but if you write the integrand as sec(x)tan(x)*sin^2(x) and integrate by parts,
you get
tan(x)sin(x) + 2cos(x)
of course this simplifies to the same thing,
but if I don't simplify it, is there any problem?
and is there a guarantee that the markers realise that they're the same thing (well maybe its 1 am when he marks this and he's really tired)
 

theVirus

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Contemporise... MAN!

ALWAYS simplify, just in case. Especially if you know that it can be simplified! You have three hours to do the examination, and 5 seconds won't hurt too much...
 

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