Complex Numbers: A Basic Manipulation

Question

$$ z = e^{i \phi} \tan \Bigg (\frac{\theta}{2} \Bigg). $$

What is,

$$ \frac{|z|}{1 + |z|^2} $$

Answer 1

Assume $\theta\space\wedge\phi\space\in\mathbb{R}$ and knowing that $\cos^2(x)+\sin^2(x)=1$

$$\left|e^{\phi i}\tan\left(\frac{\theta}{2}\right)\right|=\left|e^{\phi i}\right|\left|\tan\left(\frac{\theta}{2}\right)\right|=$$ $$\left|\cos(\phi)+\sin(\phi)i\right|\left|\tan\left(\frac{\theta}{2}\right)\right|=$$ $$\sqrt{\cos^2(\phi)+\sin^2(\phi)}\cdot\left|\tan\left(\frac{\theta}{2}\right)\right|=$$ $$1\cdot\left|\tan\left(\frac{\theta}{2}\right)\right|=\left|\tan\left(\frac{\theta}{2}\right)\right|$$

So, we get:

$$\frac{|z|}{1+|z|^2}=\frac{\left|\tan\left(\frac{\theta}{2}\right)\right|}{1+\left|\tan\left(\frac{\theta}{2}\right)\right|^2}=\frac{\left|\tan\left(\frac{\theta}{2}\right)\right|}{\left|\sec\left(\frac{\theta}{2}\right)\right|^2}$$

Answer 2

Quite simply, $\mathrm e^{i\phi}$ denotes the complex numbers of module $1$, hence $$\lvert z\rvert=\lvert\mathrm e^{i\phi}\rvert\Bigl\lvert\tan\frac\theta2\Bigr\rvert=\Bigl\lvert\tan\frac\theta2\Bigr\rvert,$$ so that $$\frac{\lvert z\rvert}{1+\lvert z\rvert^2}=\frac{\Bigl\lvert\tan\dfrac\theta2\Bigr\rvert}{1+\tan^2\dfrac\theta2}=\cos^2\frac\theta2\,\Bigl\lvert\tan\frac\theta2\Bigr\rvert=\Bigl\lvert\sin\frac\theta2\cos\frac\theta2\Bigr\rvert=\tfrac12\rvert\sin\theta\rvert.$$