Multiplicative inverse of complex numbers proof I recently attempted to show that the multiplicative inverse for complex numbers exists and expressed it in complex form, as follows:
Suppose $z = a + bi$ is a non-zero complex number. Show that $z$ has a multiplicative inverse and express it in the form $c + di$.
Let $z^{-1}$ denote the multiplicative inverse of Z. Then,
$$z^{-1}z = 1 = zz^{-1}$$
$$\implies z^{-1}(a+bi) = 1 = (a+bi)z^{-1}$$
So,
$$z^{-1} = \frac{1}{a+bi}$$
Multiplying the numerator and denominator by the conjugate:
$$z^{-1} = \frac{a-bi}{a^2 + b^2}$$
$$z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$$
Thus, for all non-zero complex numbers $z$, there exists a multiplicative inverse, $z^{-1}$, where $z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$
QED.
However, I was told that this proof is circular because I assumed that the inverse exists. How can I rectify this?
Responses are much appreciated.

What you have done is useful but it is not a proof. Now just verify that $$(a+bi)(\frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2}))$$ $$=(\frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2}))(a+bi)=1$$ by direct calculation.
