Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question
I would explain by saying that the vector OC (gamma)is OA (alpha) rotated anti-clockwise by 2pi/3 and scaled by root 2
Re: JR 2012 4U Trial 11d) Geometrical Complex Numbers Question
Yeah the part that says \gamma = \beta cis(\frac{2 \pi}{3}) is incorrect lol, skip that part.
Seeing as the person is in year 11 it's unlikely OP would know the integration basics yet, (unless OP is accelerated).
I would recommend graphs, and some polynomials, (assuming you've learned most of the theorems in 3U).
Nice paper dude :D
you have a right angled isosceles triangle, so you get x = -y (it's negative since it's below the x-axis)
Equate the function from above and make t the subject.
Yeah I don't think they would actually deduct marks for that, but if you had to compute a definite integral with negative limits, then you'd have to include the absolute value.
The only reason I'd drop to 3U from your position is that your other marks are suffering as a result of putting in too much time into 4U math, other than that, your mark's are fine, why would you even consider that?
Yes you will get E4 if you keep this up.
Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
If a quadratic can be factorised in a form of a perfect square,it would only have one root, and letting the discriminant equal to 0 implies there's only one root.
Re: UNSW chit chat thread 2016
I swear, they should give pity marks for that much error.
I remember during my complex analysis exam, there was this question that would usually take about a minute or 2 since we've seen it a lot of times, except the equation given somehow didn't work out, it...
Re: UNSW chit chat thread 2016
well SCIF is being a shit head to me, pretty sure we made the best interview with the scientist piece ever, nek minnit, 13/21..... WHAT THE FUCKKK
With inverse functions, the domain and range basically swap around, e.g
y = e^x would have a domain of all real x to have an inverse function, and the range is y>0
so the inverse function, y = ln(x) would have a domain x>0 and range all real y.