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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.

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3 Answers

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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.

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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.

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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.

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