Re: HSC 2018 MX2 Marathon
yesProve by Induction that
is this the question?
yesProve by Induction that
is this the question?
So let z = x + 2ifind z suvh that (Im)z=2 and z^2 is real
thank you
thank you Pikachu!!So let z = x + 2i
(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0
Hence z = 2i
how about if re(z) is 2Im(z) and z^2-4i is realSo let z = x + 2i
(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0
Hence z = 2i
how about if re(z) is 2Im(z) and z^2-4i is real
I cant get the answer.. the answer is +- (2+i)
Continuing on, prove that the area of the ellipse defined byProve the ellipses:
have the same area as the ellipse
Change to polar form, apply de Moivre's theorem, done. For all of them.
Realise the denominator and use De Moivre's at the end to get the n in 2nthetaThis one pls
View attachment 34383
Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)
What is the modulus and argument of z in this case? I got an answer that was different to the one in the textbook.Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.