The thing to understand is that the set of all points $(a,b)$ where $a^2 + b^2 = 1$ are precisely the points $(a,b)$ on the unit circle and they are in one to one correspondence with the real numbers $[0, 2\pi)$ so that $(a,b)=(\cos \theta, \sin \theta) \leftrightarrow m\angle AOX; A=(a,b); O=(0,0); X=(1,0)$. That should be absolute basic first week of trigonometry.
Now if $z = m+ni$ then $|z|= \sqrt{z\cdot \overline z} = (a+bi)(a-bi)=\sqrt{m^2 + n^2} =1$.
Now, claiming the trivial fact that for $z \ne 0$ we have $|z|\frac z{|z|} = z$, serves to draw attention that there is a complex number $w = \frac {z}{|z|} = \frac m{\sqrt{m^2 + n^2}} + i\frac n{\sqrt{m^2 + n^2}} = a+bi$ where $a = m{\sqrt{m^2 + n^2}}$ and $b = \frac n{\sqrt{m^2 + n^2}}$ where $|w| = 1$ and $a^2 + b^2 = 1$.
Such a complex number will always exist and $|w|$ will equal $1$ and if $w = a+bi$ then.... $|w| = \sqrt {a^2 + b^2} =1$ and therefor $a^2 + b^2 = 1$.
And by basic trig, there cooresponds a precise and unique $\theta: 0 \le \theta < 2\pi$ so that $a = \cos \theta$ and $b = sin \theta$.
This isn't profound. When I was in high school, I would have been taught that if we plotted the number $z=m+ni\ne 0$ on the Complex Plane as the point $(a,b)$ then the distance from $(a,b)$ to $(0,0)$ would be $r= \sqrt{a^2 + b^2}$ and the angle of $(a,b)$ to $(0,0)$ to the $x-axis$ would be some precise angle $\theta$, and that of this $r > 0$ and $\theta$ than $(a,b)$ would be the only such point and that $a+bi$ is the unique one and only complex number with that magnitude and with that angle.
And that's all Spivak is saying.
But what's wrong with saying that. Well, that presupposes that the "Complex Plane" makes sense and has been defined and that we can talk about complex numbers $a+bi$ as points in the plane. This is the argument that we can.