The amplitude of $1+\cos x - i\sin x$

Question

We can write as $$\cos 0 + i\sin 0 + \cos\frac x2- i\sin\frac x2$$ $$2\cos^2\frac x2-2i\sin\frac x2 \cos \frac x2$$ $$2\cos\frac x2(\cos\frac{-x}{2} + i\sin\frac{-x}{2})$$ How should I proceed

The answer is $\frac{-x}{2}$

if possible please don’t give answe in the Euler form, I want to do it without that (or please give both, that would be much more helpful)

Answer 1

We have that by $z=1+\cos\frac x2 - i\sin\frac x2$

$$|z|^2=z\cdot \bar z=(1+\cos\frac x2 - i\sin\frac x2)(1+\cos\frac x2 + i\sin\frac x2)=\\ =2+2\cos\frac x2$$

therefore

$$|z|=\sqrt{2+2\cos\frac x2}$$

The argument is

$$\arg(z)=\arctan\left(-\frac{\sin\frac x2}{1+\cos\frac x2}\right)$$

and since by trigonometric indentities

$$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} $$

we have

$$\arg(z)=\arctan\left(-\frac{\sin\frac x2}{1+\cos\frac x2}\right)=\arctan\left(-\tan \frac{x}{4}\right)=\arctan\left(\tan \left(-\frac{x}{4}\right)\right)=-\frac x 4$$

for $\frac x4\in\left(-\frac{\pi}2,\frac{\pi}2\right)$.

Answer 2

Take the steps below

$$z=1+\cos x - i\sin x$$

$$=2\cos\frac x2(\cos\frac{-x}{2} + i\sin\frac{-x}{2})= 2\cos\frac x2 e^{-i \frac x2} $$

Since the amplitude for $z=|z|e^{i\theta}$ is defined as $amp(z) =\theta$, we have,

If $\cos \frac x2 \ge 0$, $z= |2\cos\frac x2 |e^{-i \frac x2}$ and $amp(z) = -\frac x2$.

If $\cos \frac x2 < 0$, $z= |2\cos\frac x2 |e^{i\pi-i \frac x2}$, $amp(z) = \pi -\frac x2$.