How to show that $zw=1\implies w = z^{-1}, z = w^{-1}$? I need to show that
$$zw=1\implies w = z^{-1}, z = w^{-1}$$
But I don't know how to start. I've tried to consider $z=a+bi, w = c+di$ then:
$$zw = ac+adi+bci-bd\implies zw = ac-bd + (ad+bc)i = 1\implies \\ac-bd = 1\\ad+bc = 0$$
$$ac = 1+bd\\ad = -bc$$
$$c = \frac{1+bd}{a}\\d = \frac{-bc}{a}$$
$$c = \frac{1+b(\frac{-bc}{a})}{a}=\frac{1-\frac{b^2c}{a}}{a} = \frac{a-b^2c}{a^2}\implies ca^2 = a-b^2c\implies ca^2+b^2c=a\implies c[a^2+b^2] = a\implies c = \frac{a}{a^2+b^2}$$
Using $$d=\frac{-bc}{a}\implies d = \frac{-b\frac{a}{a^2+b^2}}{a} = \frac{-b}{a^2+b^2}$$
LoL I did it :D

If you know in advance that the complex numbers form a field then $$zw = 1 \implies z^{-1}zw = z^{-1}1 \implies w = z^{-1}.$$
