# Geometrically explaining $i^2 = -1$

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 $$\vert 1 \vert$$

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".
$$i$$ 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 $$i^2=-1$$.