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I never got a chance to take complex analysis in college so I decided to study it on my own. In the beginning of the books they start by proving some properties of complex numbers however I noticed that most of the proofs involve (usually implicitly) the assumption that $i/i = 1$ It is not obvious to me that this should be true so I am trying to prove it myself.

I want to prove that $i/i = 1$ starting with the definition that $i ^ {2} = -1 $ my first thought was to simply apply complex division like so.

$ \frac{0 + i}{0 + i} = \frac{0 + i}{0 + i} \left (\frac{0 - i}{0 - i} \right ) = \frac {0 - 0i +i0 - (i \cdot i)}{0 - 0i +i0 - (i \cdot i)} = \frac {1}{1} = 1$

However, it occurs to me that in multiplying by the conjugate divided by itself, I am implicitly assuming that $\frac{i}{i} = 1$ which is of course circular reasoning.

How should I approach this?

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    $\mathbb{C}$ is a field and if $x$ is a non-zero element of a field $\frac{x}{x}=1$ holds. – Jack D'Aurizio Feb 28 '17 at 20:09
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    What do you think $\frac{1}{i}$ is? – Arnaud D. Feb 28 '17 at 20:14
  • @ArnaudD. To show that $1/i=-i$ is a duplicate, implying $i/i=1$, too. – Dietrich Burde Feb 28 '17 at 20:15
  • See also this question – Ben Grossmann Feb 28 '17 at 20:17
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    I don't think this question is a duplicate of either of those. It seems to stem more from a shaky understanding of what $/i$ represents. – rschwieb Feb 28 '17 at 20:18
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    @rschwieb No, this question is not a duplicate. Arnaud asked about $1/i$, and that definitely is a duplicate. On the other hand, Jack's answer in the comment says it all. – Dietrich Burde Feb 28 '17 at 20:20
  • @DietrichBurde Oh, I misunderstood the exchange. Sorry! – rschwieb Feb 28 '17 at 20:21
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    @DietrichBurde Actually, what I meant was "What do you think $\frac{1}{i}$ is, if it is not the inverse of $i$?", as it seemed to me that the OP had not fully understood what the notation meant. – Arnaud D. Feb 28 '17 at 20:44
  • @Arnaud D I assumed it was the multiplicative inverse but at this point I am only deriving some very basic properties of the complex numbers. I have yet to show that it is a field. (not that I doubt it is, i just haven't gotten as far as proving it yet, and as my teachers were fond of saying ... you cant use it until you prove it) –  Feb 28 '17 at 20:50
  • @SE typically, you don't use the division symbol unless you're sure that you're working over a field so that division "makes sense". – Ben Grossmann Mar 01 '17 at 13:08

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You should ask yourself the following: what does "division" mean, anyway? I would say this: once you know how multiplication works, the quotient $a/b$ means the number $x$ for which $bx = a$. For example, I would say that $$ 20/4 = 5 $$ because $x = 5$ is the unique solution to $$ 4x = 20 $$ Now: what should $i/i$ be? It should be the value of $x$ for which $$ i = i\,x $$

There's an interesting technical issue here: how do we know that $x$ is the only number for which $i\,x = i$? Here's a justification I like: if we "combine like terms" we can rewrite the equation as $$ i(x - 1) = 0 $$ Now, we can use the following fact: if $ab = 0$, then either $a = 0$ or $b = 0$. Since $i \neq 0$, it must be the case that $(x-1) = 0$, which is to say that $x = 1$.

Of course, the statement "if $ab = 0$, then either $a = 0$ or $b = 0$" should be proven. However, I find it very intuitively believable, so I'm content to leave it at that.

Ben Grossmann
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    It is "intuitively obvious" that $\mathbb R$ is an integral domain -- but at least to me much less so that $\mathbb C$ is. (The only way I can think of proving it right away is to show that all nonzero elements have multiplicative inverses -- and then we might, for this case, just as well multiply $i=ix$ by $-i$ on both sides, yielding $1=x$). – hmakholm left over Monica Feb 28 '17 at 21:17
  • @HenningMakholm I would note that $|wz| = |w| , |z|$ for complex numbers $w,z$. – Ben Grossmann Feb 28 '17 at 21:58
  • hmm, yes, that would work too. – hmakholm left over Monica Feb 28 '17 at 21:58
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The notation $x/y$ is just another way of writing $xy^{-1}$. By definition, $yy^{-1}=y^{-1}y=1$. So $i/i=ii^{-1}=1$.

This all follows from the definition of inverses and the notation we use for them. There is nothing to prove, really.

rschwieb
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Your proof would be as simple as citing an axiom.

$i * 1 = i$ because 1 is the multiplicative identity.

dividing both sides of the equation by $i$ gives us:

$1= \frac{i}{i}$

Eric Haney
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Think of complex multiplication as rotation and scale of the number being multiplied.

Every real number number multiplied by $i$ rotates $90\deg$ counterclockwise in the complex plane. Because division is just reversed multiplication, dividing by $i$ rotates the number clockwise. When you start with $i$ the point $0+i$ lays in the $y$-plane of the complex plane $1$ units of distance from the origin, if you rotate it clockwise you get the correct answer on the real line, $1$.

Garmekain
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You can see the identity in this way: $$\displaystyle\frac{1}{i}=-i,$$

in this way you obtain $-i\cdot i=-(-1)=1.$

InsideOut
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