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.
 A: 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.
A: Just show that the "expanded" form of the inverse satisfies the properties of the multiplicative inverse without assuming the existence of the inverse itself, this avoids circularity. Finding that $$\frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2}$$ is the "scratch work" of the proof that you don't actually show when writing it up formally.
A: there's a few things wrong.  One very subtle.
First of all we have utterly no idea what a number of the form $\frac 1{a+bi}$ even means.  All we did was write a $1$ put a bar underneath it and write $a+bi$ under that.  We can make up rules that somehow $\frac {a+bi}{c+di}$ (whatever that means) when multiplied but  $\frac {e+fi}{g+hi}$ will be equal to $\frac {(a+bi)(e+fi)}{(c+di)(g+hi)}$ but that doesn't mean anything.
We have to define that $\frac 1z$ must mean a complex number $w$ so that $z \cdot w = 1$ (assuming that there is such a number, and that it is unique; neither of which we have any reason to assume).  And even if we do assume there is a $w$ so that $w(a+bi) =1$ and we write it as $w=\frac 1{a+bi}$ and if there is $v$ so that $v(c+di) = 1$ so we can write $v=\frac 1{c+di}$ we have no reason to believe that $wv = \frac 1{a+bi}\cdot  \frac 1{c+di}$ that that will actually equal $\frac 1{(a+bi)(c+di)}$.  (Although we can prove that.)
Anyhoo.....
So long as $z=a+bi = 0 \iff a^2 + b^2 = 0$ then the does exist a $w= \frac a{a^2 + b^2} -i\frac b{a^2 + b^2}$ and it is true that $(a+bi)(\frac a{a^2 + b^2} -i\frac b{a^2 + b^2}) = (a\cdot \frac 1{a^2+b^2} + b\frac 1{a^2 + b^2}) + i(a \frac b{a^2+b^2} - b\frac a{a^2+b^2}) = \frac {a^2 + b^2}{a^2 + b^2} + i(\frac {ab}{a^2+b^2} - \frac {ab}{a^2 + b^2}) = 1$
So an inverse does exist.  But we must also prove it is unique.  Now the way I'd do it, I'd simbly set up an equation $(a+bi)(c+di) =1$ and solve for $c$ and $d$ and show the solution is unique.... but  it's a little too late for that!
I'd say though if $(a+bi)(c+di) =1$ and $(a+bi)(e+fi)=1$ then $c+di = (c+di)\cdot 1 = (c+di)\cdot (a+bi)(e+fi)= [(c+di)(a+bi)](e+fi)= 1(e+fi) = e+fi$.  So there is only one possible solution and we know $\frac a{a^2 + b^2} -i\frac b{a^2 + b^2}$ is one solution, so it is the only solution.
