Question: I was asked to show the following equality:
$$|1-z|=2\sin \Big( \frac{\arg z}{2} \Big) $$
where $|z| = 1$. It is not too hard to prove it. It comes out naturally from doing some computations. However I am having a hard time understanding the geometric meaning. I was wondering if anyone here knows why this is correct geometrically / intuitively.

Thank you!

Answer: 
[![enter image description here](https://i.stack.imgur.com/zOnKN.png)](https://i.stack.imgur.com/zOnKN.png)

In the argand plane, the point $B$ on the circumfurence of the unit circle represents the complex number $z$ with $|z|=1$. $C$ represents the point $1$. We see that $$|1-z|=|BC|.$$ $AD$ is the angle bisector of $\angle BAC$. $\Delta BAC$ is isosceles, as $|AB|=|AC|=1$. This implies $$|BD|=|DC|=\frac12|BC|=\frac12|1-z|.$$ Also $AD\bot BC$, hence $$\sin(\arg(z)/2)=\sin(\angle DAC)=\frac{|CD|}{|AC|}=|CD|.$$
Draw the triangle with vertices $A=1$, $B=0$, and $C=\frac{z-1}{2}$. This is a right triangle with angle opposite $AC$ equal to $\frac{\arg z}{2}$. It follows from the definition of $\sin$ that $\sin(\frac{\arg z}{2})=|AC|=\frac{|1-z|}{2}$