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Hi godzilla, The problem is with your "We know that" step. That integral \int_{-\infty}^{+\infty} f(x)\ dx is, so to speak, "designed" to be one; its value was known to be the value described above and then it was divided by a constant to make it equal to one. This is called "normalizing" (not...

I don't know if it helps anybody, but in that question you wrote for part (ii) "Show that w conjugate is also a root". In part (i) you proved (hopefully) that w is a root, thus there is really nothing to do for part (ii). Since the coefficients of the polynomial are all real, it is known that...

In case you still need some help, here are some ideas for the second case (also applicable to the first case, as someone has already hinted): You can solve as follows: \because \frac{\alpha}{\beta \gamma} = \frac{\alpha^{2}}{\alpha \beta \gamma} = \alpha^{2} it follows by symmetry that the...

This step is not correct; in writing z = x+ iy we assume y to be a real number, therefore y^2 = -3 has no solutions. You can easily verify this by substituting 1 + iroot3 into the original equation. Your first two solutions were correct, of course.

For by far the coolest (live fast, die young) life: Evariste Galois. For prolific-ness (if that's a word): Karl Friedrich Gauss For most likely to have been not a human: S. Ramanujan (don't ask me to spell his first name) For sheer overall insanity: Georg Cantor (but there's a lot of...

On the approach: Use the known relationships for modulus and argument of product and quotient: arg(z_1/z_2) = arg(z_1) - arg(z_2) (+2*k*pi, if you must...) |z_1/z_2| = |z_1|/|z_2| Thus in the case the difference in the arguments of z-w and z+w is equal to the argument of 2i, which is pi/2...

I notice nobody mentioned the incredibly beautiful solution to the problem of integrating this function between -infinity and plus infinity. Since there is no latex here, I won't write it, but you can read a nice walkthrough here: http://mathworld.wolfram.com/GaussianIntegral.html It's called...

Actually -1 mark for writing +k3 instead of -k3 at the end there :)

I have a curious way of answering this:

Here's my answer (with pictures :) ) 3unitz, I like your answer, if anything I prefer it, at least for school level. Even though you made an error, you can rejig it to find the other side of the triangle and it's right. It's always nice to have both an algebraic and a geometric way to see an...

3unitz: You seem to be finding the length of e^6ix + e^-5ix instead of e^6ix - e^-5ix. That's why your final answer is cosines (using sin 2@ formula) instead of sines.

Thanks Affinity, that's the symmetry I was talking about :) I have a pretty picture of the answer, if anyone wants, I can post it later. 3unitz - interesting, I'm pondering your attempt - trying to work out your sine rule business there ...

Up to here, perfect! :) But the next step is a mistake. Actually that's the step that isn't so obvious. Look for a hidden symmetry. About Euler, i.e. ei@=cis@ : You're right! I teach British Maths A-level syllabus, we let that one slide in a lot earlier in our teaching, I just rather...

Attached as an image. Comments: If you find yourself doing long drawn out trig algebra, you might get to the answer, but you're going the wrong way. The geometrical interpretation of the answer is particularly interesting (well, to me anyway).

Would that be the American SATs? Or is there an Australian one?

Hi, I enter the forum with some trepidation as I'm neither an Australian nor even a student - so forgive me barging in here. I'm teaching A-level Maths in China (A-level is the British exam in case you don't know). We have a lot of very talented and able students here. We've been trying to find...