Geometrically explaining $i^2 = -1$

Question

My textbook, A first course in Abstract Algebra by Fraleigh, 7th edition, (pg14) is attempting to explain complex number multiplication geometrically.

The math example walks through multiplying $z_1 = \vert z_1 \vert e^{i\theta_1} $and $ z_2 = \vert z_2 \vert e^{i\theta_2}$ into

$$z_1 z_2= \vert z_1 \vert \vert z_2 \vert \bigg(cos(\theta_1 + \theta_2) + isin(\theta_1 + \theta_2 \bigg)$$

This makes sense so far. The book concludes with the following (paraphrasing for space):

We multiply complex numbers by multiplying their absolute values and adding their polar angles.

If $i$ has polar angle $\frac{\pi}{2}$ and $\vert 1 \vert$, then $i^2$ has polar angle $2\frac{\pi}{2} = \pi$ and $\vert 1 * 1 \vert = 1$, so that $\textbf{$i^2$ = -1}$

This last part confuses me, I understand that $i^2$ is supposed to = $-1$, but I don't see where they draw the connection given $i = $ $\frac{\pi}{2}$ and $\vert 1 \vert$

Answer 1

Let $z_1=z_2=i$. Note that $$ i=|i|e^{i\frac{\pi}{2}}=e^{i\frac{\pi}{2}}. $$ Let $\theta_1=\theta_2=\pi/2$. Now use the formula that "makes sense so far".

Answer 2

$i$ can be represented by $(r,\theta) = (1,\pi/2)$ with polar coordinates. This is a point on the imaginary axis. When you have $i^2$, this can be represented as $(1,\pi)$ which is back on the real axis. Because this is left of the origin, it is the negative part of the real axis and thus you have $i^2=-1$.

Please see here for diagrams and formulas for transforming to and from polar coordinates: https://en.wikipedia.org/wiki/Polar_coordinate_system#Complex_numbers