<s>[INST] 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
 [/INST] Hint Multiply both $zw=1$ by $z^{-1}$. Do the same for $w$.
Also, how do you define $z^{-1}$?
Since complex multiplication is commutative, the definition of $z^{-1}=w$ is $zw=1$. So there is absolutely nothing to prove.... 
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