Find modulus and argument of $\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}}$

Question

A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.

Given that:

$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}} $$

To prove:

$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = \theta $$

My work so far;

Proof:

$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$

I'm not sure about the above, if there's an easy explanation or method to go through would be great as questions like these are giving me a bit of trouble.

Answer 1

Hint:$$\zeta=\frac{2(\sin2\theta)(\sin2\theta+i\cos2\theta)}{2(\cos\theta)(\sin\theta+i\cos\theta)}=\frac{2\sin 2\theta\cdot ie^{-2i\theta}}{2\cos\theta\cdot ie^{-i\theta}}=2\sin\theta e^{-i\theta}.$$

Answer 2

Method 1: Using the $1-\cos2A=2\sin^2A$ identity

For $|\zeta| = 2\sin(\theta)$:

$$\zeta = {\frac {|1-\cos4(\theta) + i \sin4(\theta)|} {|\sin2(\theta)+ 2i \cos^2(\theta)|}} $$

$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$

$$\zeta = {\frac {\sqrt{(1-2\cos4\theta + \cos^24\theta+\sin^24\theta}} {\sqrt{4\sin^2\theta\cos^2\theta+ 4\cos^4(\theta})}} $$

$$\zeta = {\frac {\sqrt{(2-2\cos4\theta}} {\sqrt{4\cos^2\theta(\sin^2\theta+ \cos^2(\theta}))}} $$

$$\zeta = {\frac {\sqrt{2}\sqrt{1-\cos4\theta}} {2\cos\theta}} \begin{cases} \mbox{$1-\cos2A=2\sin^2A$ identity} \end{cases} $$

$$\zeta = {\frac {\sqrt{2}\sqrt{2\sin^22\theta}} {2\cos\theta}} $$

$$\zeta = {\frac {2\sin2\theta} {2\cos\theta}} $$

$$\zeta = {\frac {2\sin\theta\cos\theta} {\cos\theta}} = 2\sin\theta.$$

For $\arg(\zeta) = \theta$:

$$\arg(\zeta) = \arg(1-\cos4\theta+i\sin4\theta)-\arg(\sin2\theta+2i\cos^2\theta)$$

$$\arg(\zeta) = \arctan(\frac{\sin4\theta}{1-\cos4\theta})-\arctan(\frac{2\cos^2\theta}{\sin2\theta})$$

$$\arg(\zeta) = \arctan(\frac{2\sin2\theta\cos2\theta}{2\sin^22\theta})-\arctan(\frac{2\cos\theta\cos\theta}{2\sin\theta\cos\theta})$$

$$\arg(\zeta) = \arctan(\cot2\theta)-\arctan(\cot\theta)$$

$$\arg(\zeta) = \arctan[\tan(90°-2\theta)]-\arctan[\tan(90°-\theta)])$$

$$\arg(\zeta) = 90°-2\theta-90°+\theta$$

$$\arg(\zeta) = -\theta.$$

Method 2: Using De Moivre's theorem,

$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i \cos^2(\theta)}} $$

$$\zeta = {\frac {2\sin^22\theta + i 2\sin2\theta\cos2\theta} {2\sin\theta\cos\theta+ 2i \cos^2\theta}} $$

$$\zeta = {\frac {2\sin2\theta(\sin2\theta + i\cos2\theta)} {2\cos\theta(\sin\theta+ i \cos\theta)}} $$

$$\zeta = {\frac {2\sin\theta\cos\theta(\sin2\theta + i\cos2\theta)} {\cos\theta(\sin\theta+ i \cos\theta)}} * {\frac{-i}{-i}}$$

$$\zeta = {\frac {2\sin\theta(\cos2\theta - i\sin2\theta)} {\cos\theta - i \sin\theta}}$$

$$\zeta = {\frac {2\sin\theta[(\cos-2\theta + i\sin-2\theta)]} {\cos(-\theta) + i \sin(-\theta)}}$$

$$\zeta = {\frac {2\sin\theta(\cos\theta+i\sin\theta)^{-2}} {(\cos\theta+i\sin\theta)^{-1}}} \begin{cases} \mbox{De Moivre's Theorem} \end{cases} $$

$$\zeta = 2\sin\theta[\cos(-\theta)+i\sin(-\theta)]$$

$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = -\theta $$

Answer 3

Note that\begin{align}\bigl(1-\cos(4\theta)\bigr)^2+\sin^2(4\theta)&=2-2\cos(4\theta)\\&=2-2\bigl(1-2\sin^2(2\theta)\bigr)\\&=4\sin^2(2\theta)\\&=16\sin^2(\theta)\cos^2(\theta)\end{align}and that\begin{align}\sin^2(2\theta)+4\cos^4(\theta)&=4\sin^2(\theta)\cos^2(\theta)+4\bigl(1-\sin^2(\theta)\bigr)^2\\&=4\sin^2(\theta)-4\sin^4(\theta)+4-8\sin^2(\theta)+4\sin^4(\theta)\\&=4\bigl(1-\sin^2(\theta)\bigr)\\&=4\cos^2(\theta)\end{align}and that therefore$$\lvert\zeta\rvert^2=\frac{\bigl(1-\cos(4\theta)\bigr)^2+\sin^2(4\theta)}{\sin^2(2\theta)+4\cos^4(\theta)}=4\sin^2(\theta).$$