modus

Despite Melody saying she didn't have time to write up and check her answer, what she wrote was nevertheless almost complete. The only omission was in the "Let M be the midpoint of AC" bit. For the sake of those who want a complete proof this bit should say:–

"Since AD is parallel to BC we must have CAD = ACB (as indicated in the diagram, but without justification). Drop a perpendicular BM from B to AC. At this stage we dont know that M is the midpoint of AC. But the two triangles AMB, CMB are congruent (angle, angle, side). Therefore AM=MB."

The second "proof" isn't really a proof at all, because it assumes that \(\alpha=45^\circ\), which isn't necessary at all, as Melody has shown.

By the way, we've never heard of a trapezoid in Mathematics in the UK (including England) either. It is the universal name of a bone in your hand, so if you were up in Anatomy you would have heard of it :-) The only name in the UK for a 4-sided figure with no parallel sides is a quadrilateral (admittedly an irregular one).

Guest, your last 4 answers are correct!

To compute: \(\displaystyle\int^{\infty}_{0}\dfrac{\ln^2\left(z\right)dz}{z^4+z^2+1}\).

I think this does come out by Cauchy, but it's quite heavy.

First you need the poles created by the zeros of \(z^4+z^2+1=0\): first put \(Z=z^2\) so \(Z^2+Z+1=0.\) The roots of this equation are the complex cube roots of unity \(\omega_1={1\over2}+i\sqrt{3\over2}\) and \(\omega_2=\overline\omega_1=\omega_1^2\).

Therefore the roots of \(z^4+z^2+1=0\) are the square roots of \(\omega_1,\ \omega_2\). But since \(\omega_2^2=\omega_1\)and \(\omega_1^2=\omega_2\) these four square roots are just \(z=\pm\omega_2\) and \(z=\pm\omega_1\).

You then have to compute the residues at these 4 poles: it helps that \(\omega_1=\exp(2\pi i/3)\) being a cube root of unity has absolute value 1 which simplifies the computation of \(\ln(z)\).

Having got the residues you then do the usual contour integration trick round two circles radius R and r joined by a cut up the negative real axis. This very much follows the example given in Wikipedia for the case where the denominator is just \((1+z^2)^2\) given here: