How can I show the following trigonometric function

Question

How can I show that $$|\frac{18k(a^{4}-z^{4})}{82a^{2}z^{3}-9z(a^{4}+z^{4})}|=\frac{18k\sin(2\theta)}{a(41-9\cos(2\theta))}$$

Where $z=ae^{i\theta}$, this the point I've gotten to in a calculation and I'm Stuck here , how would I calculate the modulus , and do I need to rationalize the denominator and if so how would I find the complex conjugate of the bottom?

Answer 1

$$|\frac{18k(a^{4}-z^{4})}{82a^{2}z^{3}-9z(a^{4}+z^{4})}|=|\frac{18ka^4(1-e^{i4\theta})}{82a^{5}e^{i3\theta}-9a^5(\cos\theta+i\sin\theta)(1+e^{i4\theta})}|$$

$$= |\frac{18k(1-\cos 4\theta-i\sin4\theta)}{82ae^{i3\theta}-9ae^{i\theta}(1+\cos 4\theta+i\sin4\theta)}|$$

$$= |\frac{36k\sin2\theta(\sin2\theta-i\cos2\theta)}{82ae^{i3\theta}-18ae^{i\theta}(\cos2\theta+i\sin2\theta)\cos2\theta }|$$

$$= \frac{36k\sin2\theta *|e^{-i(\frac{\pi}{2}-2\theta)}|}{(82a-18a\cos2\theta)|e^{i3\theta}| }$$

Since $|e^{ix}|=1,$

$$= \frac{18k\sin(2\theta)}{a(41-9\cos(2\theta))}$$