# Why is $i/i$ equal to 1

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?

• $\mathbb{C}$ 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
• What do you think $\frac{1}{i}$ is? – Arnaud D. Feb 28 '17 at 20:14
• 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
• @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 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

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.

• 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

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.

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:

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

• When you "divide both sides by $i$", you assume that $$\frac{i}{i} * 1 = 1$$ – Ben Grossmann Feb 28 '17 at 20:22
• I am assuming the existence of inverses, another of the field axioms. The comment by Jack D'Aurizio and rschwieb address this more directly. – Eric Haney Feb 28 '17 at 20:32

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, $1$.

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

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