We have developed interesting properties of adding vectors with equal magnitudes. Similar properties can be written in terms of complex numbers. Let $z_1$ and $z_2$ be complex numbers with $\left|z_1\right|=\left|z_2\right|=r$. Set $z_1=r \operatorname{cis} \theta_1$ and $z_2=r \operatorname{cis} \theta_2$. Set $z=z_1+z_2$. Then $O Z_1 Z Z_2$ is a rhombus, implying that the line $O Z$ bisects the angle $Z_1 O Z_2$; that is, $z=r^{\prime}$ cis $\frac{\theta_1+\theta_2}{2}$ for some real number $r^{\prime}$. In particular, if $z_1=1$ and $z_2=\operatorname{cis} a$, then $z=r^{\prime}$ cis $\frac{a}{2}$, where $r^{\prime}=\overline{O Z}$. Therefore, $\tan \frac{a}{2}$ is equal to the slope of line $O Z$; that is, $$ \tan \frac{a}{2}=\frac{\sin a}{1+\cos a}, $$ which is one of half-angle formulas. Other versions of the half-angle formulas can be obtained in a similar fashion. It is also not difficult to see that $r^{\prime}=\overline{O Z}=2 \cos \frac{a}{2}$.
The previous paragraph is taken from 103 trig problems, the idea is to extend the notion of properties found in normal vectors to ones that use complex numbers. But in this case, they say it should be without difficulty to see that $ 2\cos \frac{a}{2}$ = $r^{\prime}$ but I have tried to look for a solution and have been unable to find how it was derived.
It's $2\cos\frac a2$, not $2\cos a$. Look at the rhombus. The angle between the diagonal of length $r'$ and the horizontal axis is $\frac a2$, since the diagonal is also bisector. From $(1,0)$ draw a perpendicular to the diagonal (this will also be the other diagonal of the rhombus), say to point $M$. Then $\triangle OMZ_1$ and $\triangle ZMZ_1$ are congruent (right angle triangles, the hypotenuse is $1$, and $MZ_1$ is common). That means $$OM=\frac12OZ=\frac12r'$$ and $$OM=OZ_1\cos\frac a2=\cos\frac a2$$ Therefore $$r'=2\cos\frac a2$$