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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.

SupremePickle
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    You have proven that if an inverse of $a+bi$ exist, it must be equal to $\frac a{a^2 +b^2} - i\frac{b}{a^2+b^2}$. But you haven't proven it exists. So add the one line by taking pen to paper and writing "And so long as $a^2+b^2 \ne 0$ then $\frac a{a^2+b^2}$ and $\frac b{a^2+b^2}$ are real numbers and such an inverse does exist". ANd then notice that "But no such number does exist if $a^2 + b^2=0$. $a^2 + b^2=0$ for real $a,b$ if and only if $a=b=0$ so $z=0$ is the only complex number without an inverse". (You do have to note $z=0$ doesn't have an inverse. – fleablood Oct 20 '20 at 00:18
  • @fleablood I would add that you should verify that the number you found is the inverse of $a+bi$. The work so far shows that this is the only possible candidate; all that is left to do is to verify that it works (there might be no inverse at all, a priori). – Servaes Oct 20 '20 at 08:19
  • Another way to go about it, is to show that all implications ($\Rightarrow$) in the proof are in fact equivalences ($\Leftrightarrow$), if $z\neq0$. – Servaes Oct 20 '20 at 08:20
  • Yeah, after thought I think my comment is circular. .....We can multiplyb by $\frac {a-bi}{a-bi}=1$ unless we assume all complex numbers have inverses. I think the best way to prove a unigue inverse exist but only for non-zero numbers is to solve $(a+bi)(c+di)=(ac-bd)+(bc+ad)i=1$ has only solution $c=\frac a{a^2+b^2};d=\frac b{a^2+b^2}$. – fleablood Oct 20 '20 at 15:24

3 Answers3

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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.

Kavi Rama Murthy
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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.

Rough L
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  • So do I just plug in the resulting $\frac{a-bi }{a^2 + b^2}$ into the $zz^{-1} = 1$ at the end of the above? – SupremePickle Oct 20 '20 at 00:11
  • Recall that the inverse of an element $z$ is the unique element $z^{-1}$ such that $zz^{-1} = z^{-1}z = 1.$ You just have to show that given any $a + bi \in \mathbb{C}$, that $\frac{a - bi}{a^2 + b^2}$ satisfies this properties, then by definition it is the inverse. – Rough L Oct 20 '20 at 00:15
  • Gotcha. But would I have to remove anything from what I wrote above? – SupremePickle Oct 20 '20 at 00:16
  • Yes. By assuming that $\frac{1}{a + bi}$ is the inverse and then showing that it is equivalent to that other form, you are using circular reasoning. Just start with any element of the form $a + bi$ and show that the inverse you found satisfies the inverse property. – Rough L Oct 20 '20 at 00:17
  • Sorry but I'm a little lost by what you mean. Looking back at what I wrote, I assumed that an inverse exists and showed that the inverse equals to $\frac{a-bi}{a^2 + b^2}$. So what exactly are you suggesting? Are you just saying that I should check if $\frac{a-bi}{a^2 + b^2}$ satisfies the multiplicative inverse property after what I wrote above? Or what exactly should I remove from above? – SupremePickle Oct 20 '20 at 00:25
  • You cannot start a proof of the existence of some object by assuming the existence of that object. That is why it is a circular argument: you are assuming that the inverse exists in order to prove that the inverse exists. The proof is not deriving the expression for the inverse, but rather showing that said expression satisfies the definition of the inverse, thus proving the inverse's existence. – Rough L Oct 20 '20 at 00:30
  • Ok I see. So how do I derive the expression without assuming the inverse's existence? – SupremePickle Oct 20 '20 at 00:32
  • I think that you misunderstand the point of the proof. It is not to derive the expression of the inverse, but rather to show that the thing that you found IS the inverse. For example, in the field $\mathbb{R}$, the inverse of any nonzero number $a$ is clearly $\frac{1}{a}.$ To find the inverse of $a,$ we just divide 1 by $a$. But this procedure is valid because we already know that $\mathbb{R}$ is a field and has inverses. When proving the existence of the inverse, we cannot assume that the inverse exists. Once we show it does exist and in fact equals $\frac{a - bi}{a^2 + b^2}$, then we – Rough L Oct 20 '20 at 01:59
  • can freely use that expression when finding any general inverse. But the problem is not to calculate the inverse, it is to show that it exists. Think again about the inverse for some $a \in \mathbb{R}$. We don't know off the bat that $a^{-1} = \frac{1}{a}.$ The proof for the existence of the inverse of $a$ would go: let $a \in \mathbb{R}.$ Then since $\frac{1}{a} \in \mathbb{R}$ satisfies $\frac{1}{a} \cdot a = a \cdot \frac{1}{a} = 1,$ we have that $\frac{1}{a} = a^{- 1}$ in $\mathbb{R}.$ Does that make more sense? You follow this outline for the proof of the existence of the inverse in C. – Rough L Oct 20 '20 at 02:05
  • I understand now! Thank you for all the responses! – SupremePickle Oct 20 '20 at 07:21
  • You're very welcome! – Rough L Oct 21 '20 at 01:19
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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.

fleablood
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