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z^5-1=0\\\\z^5=1\\\\$Let $z={\cos\theta} + i{\sin\theta}\\\\$Using De Moivre's Theorem: $z^5=({\cos\theta}+ i{\sin\theta})^5={\cos5\theta}+ i{\sin5\theta}\\\\$and we know that $z^5=1\:$and$\ \cos\2k\pi + i\sin\2k\pi$ where k is an integer, so$\\\\{\cos5\theta}+ i{\sin5\theta}=\cos\2k\pi +...

is (z-1)(1+z+...z^n-1) meant to be a clue? not the answer?

woops...yeh i skipped that explanation, thanks for clearing that up:)

\frac{1-z^{n}}{1-z}\\\\=\frac{1-\cos n\theta-i\sin n\theta}{1-\cos \theta - i \sin...

I went to school with thisiswar and trust me she will be a good tutor for chemistry. I have bombarded her with lots of chem questions especially for HSC and she has been very patient with me and has given thorough answers which has helped in my understanding:)

ok so choose a point on the parabola x^2=4ay: P(2ap,ap^2). Now let's find the normal at that point. y=x^2/4a dy/dx=2x/4a x=2ap, dy/dx=p so gradient of normal is -1/p y-ap^2=-1/p (x-2ap) Sub (0, Ka) into the equation[since you are given that the normals pass through (0,Ka)] Ka-ap^2=...

Are you willing to travel to Auburn? http://community.boredofstudies.org/showthread.php?t=270679

don't i get any thanks? :(

ok i'll answer the first one: let the height be h, the length be 2b and the breadth be b. Using these dimensions, you will find that the surface area is 6bh + 4b^2=48. Now, from here, make h the subject. [It should be in terms of b] Now i'm assuming what you mean by "maximum value" you mean...

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CLOSED: NOT TAKING ANYMORE STUDENTS Hi there, I graduated from St George Girls High School (a top 20 selective school) with: - Top 10 internal ranks in Mathematics Extension 2 - 2nd in Year 11 Mathematics Extension 1 - 1st in Year 10 Mathematics. I just finished my HSC and received...