Prove that such an inverse is unique Given $z$ is a non zero complex number, we define a new complex number $z^{-1}$ , called $z$ inverse to have the property that $z\cdot z^{-1} = 1$ 
$z^{-1}$  is also often written as $1/z$

Whenever you need to prove the uniqueness of an element that holds some property, you can begin your proof by assuming the existence of at least two such elements that hold this property, say $x$ and $y$, and showing that under this assumption, it turns out $x = y$, necessarily.
In this case, the property we'll check is "being an inverse of $z$": We'll use the definition of $z$-inverse (the inverse of $z$): it is "any" element $z'$ such that $$z' z = zz' = 1\tag{1}$$ (We won't denote any such element by $z^{-1}$ yet, because we have to first rule out the possibility that such an element is not unique.)

So suppose $z \neq 0 \in \mathbb C$ has two inverses, $x, y,\;\;x\neq y$. We won't call either of them $z^{-1}$ at this point because we are assuming they are distinct, and that they are both inverses satisfying $(1)$.
Then we use the definition of an inverse element, $(1)$, which must hold for both $x, y$. Then


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*Since $y$ is an inverse of $z$, we must have, by $(1)$ that $\color{blue}{\bf yz} = zy = \color{blue}{\bf 1}$, and 

*Since $x$ is an inverse of $z$, we must have that $xz = \color{blue}{\bf zx = 1}$, again, so it satisfies the definition of an inverse element given in $(1)$. 
This means that $${\bf{x}} = \color{blue}{\bf 1} \cdot x = \color{blue}{\bf(yz)}x = y(zx) =\;y\color{blue}{\bf (zx)} = y \cdot \color{blue}{\bf 1} = {\bf {y}}$$
Hence, $$\text{Therefore,}\quad x \;=\; y \;= \;z^{-1},$$ 
and thus, there really is only one multiplicative inverse of $z;\;$ that is, the inverse of a given complex $z$ must be unique, and we denote it by $\,z^{-1}.$
