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Prove the equation of a line and a circle in complex plane has a general form of : \alpha \ z\overline{z}+\beta \ z + \overline{\beta\ z}+ \gamma = 0 where \alpha,\gamma \in \mathbb{R}, \beta \in \mathbb{C} Hence, or otherwise, prove If z,z_{1},z_{2} are complex numbers which...

If alb,c are sides a triangle, then prove 3(ab+bc+ca) \leq (a+b+c)^{2} \leq 4(ab+bc+ca)

Let S_{n} = \frac{5}{9} \ \frac{14}{20} \ \frac{27}{35} \ ... \ \frac{2n^{2}-n-1}{2n^{2}+n-1} then find \lim_{n\to\infty}S_{n}

Prove for n\geq 2 tan(\alpha) \ tan(2\alpha)+ tan(2\alpha) \ tan(3\alpha)+ ...+ tan((n-1)\alpha) \ tan(n\alpha) = \frac{tan(n\alpha)}{tan(\alpha)}-n

If 0< x <\frac{\pi}{2} prove sin(x) > x - \frac{x^{3}}{4}

Suppose A= 2^{n}\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2+2cos\alpha}}}}}_{n \ times} then simplify A in terms of n and \alpha , then find lim_{n\to\infty}A

If log_{a}16 + log_{\sqrt{2}}a = 9 then find a.

Suppose P(x) is a polynomial which satisfies the following condition: P(P'(x)) = 27x^{6}-27x^{4}+6x^{2}+2. find a possible polynomial, P(x), that satisfies the above condition.

Suppose a_{1},a_{2},...,a_{n} is an arithmetic sequence.Then prove \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} =\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}

This is the question: let P = (1-tan^{2}\frac{a}{2})(1-tan^{2}\frac{a}{2^{2}}) (1-tan^{2}\frac{a}{2^{3}})...(1-tan^{2}\frac{a}{2^{m}}) and by simplifying P, find lim_{m\to\infty}P

Re: HSC 2017 Maths (Advanced) Marathon To solve it in a different way, we can look at the roots, since x^{3}-8 is a cubic polynomial then there are three roots, \alpha, \beta, \gamma .Obviously one of the roots is 2,so let \alpha = 2 then sum and product of the roots imply \beta+...

The solution is correct. But your solution is exactly the same as the other one: In the textbook's solution, they subtracted two angles one at a time, meaning: \angle ABC = 180 - \angle ACB - \angle CBA But you subtracted the sum in one go, meaning: \angle ABC = 180 - (\angle ACB + \angle CBA)

a) 1 < m \leq \frac{\pi}{2} because sin^{-1}x is an odd function. So to find all possible lines we have to make sure that it passes through the origin, which it always does, and one point on the first quadrant (x,y). and by oddness of sin^{-1}x , the line also passes through (-x,-y) as...

I thought to share another method to tackle the question: \frac{1}{4}+ \frac{3}{4}cos^{2}2x = \frac{1}{4}+ \frac{3}{4}(1-sin^{2}2x) = 1 -3sin^{2}x \ cos^{2}x = 1-3sin^{2}x \ cos^{2}x \ (sin^{2}x \ + cos^{2}x) = 1-3sin^{4}x\ cos^{2}x -3sin^{2}x \ cos^{4}x = sin^{6}x \ +cos^{6}x The last...

I have experienced that in physics. I came second in the first test but in the second exam/assessment I made a huge mistake which made me come 22nd out 40 in my cohort.I was determined to close the gap and in the next exam I came in the 10th place. Finally by coming 1st in my trial, I achieved...

Another slightly different method is to multiply the original equation by (1-i) then equation only the imaginary parts yields: -5y=-10 and equating the real part of the original part shows x = 4

Expanding on that idea, we can prove in a different way why \sqrt{-6 } \sqrt{-6} = 6 is not correct. let f(x) be our extended square root function, principal square root, where (f(-1))^{2} = -1, the definition of i,. we want to prove any function with that condition can not satisfy f(x)f(y)...

A different method is to play with sum and product of the roots. let p(x) = ax^{4}+bx^{3}+x^{2}+4x+2 then from the info in the question we know p(x)-(2x+3) is divisible by Q(x)= 2x^{2}+2x-1 Note the roots of Q(x) are \frac{-1\pm \sqrt{3}}{2} let's called the other two \alpha ...

the answers are correct, as far as I can tell, but this is a proper proof: after 1 year: owed money= (20000-3000)*1.06 after 2 years: owed money= ( (20000-3000)*1.06 -3000)*1.06 = 20000*1.06^{2} - 3000*(1.06+1.06^{2}) . . . . after n years: owed money = 20000*1.06^{n} -...

Thank you for pointing that out, I fixed it.