C*-algebra and multiplying by complex numbers According to a page on C*-algebra, https://www.wikipedia.com/en/C*-algebra :
$$(x y)^* = y^* x^*$$
and for every complex number λ in C and every x in A:
$$(\lambda x)^* = \bar{\lambda} x^*$$
The bar symbolizes a conjugate. Should the second relationship be:
$$(\lambda x)^* = x^* \bar{\lambda}$$
The lambda looks like a "simple y", and thus should go on the other of x. It could be the case that the authors assume a complex number $\lambda$ commutes with $x$ so $\bar{\lambda} x^* == x^* \bar{\lambda}$. Does the notation suggest there is a difference between a bar and a conjugate?
 A: A $C^*$-alebra is by definition a $\mathbb{C}$-algebra, and thus in particular a $\mathbb{C}$-vector space. This means that there is a scalar action of $\mathbb{C}$ on the algebra $A$: you can define $\lambda\cdot x$ for any $\lambda\in \mathbb{C}$ and any $x\in A$.
But the expression $x\cdot \lambda$ is usually not really defined. Or, if we really wanted to give it meaning, it would be, by definition, $\lambda\cdot x$. This is because $\lambda$ is not an element of $A$, so the expression $\lambda\cdot x$ does not come from the multiplication $A\times A\to A$, but from the external multiplication $\mathbb{C}\times A\to A$. In particular, $\lambda^*$ does not really make sense in general (unlike $\bar{\lambda}$).
Of course, the situation is a little different when $A$ is a unital algebra (ie when it has a multiplicative unit $1_A$). In that case, the elements of the form $\lambda\cdot 1_A$ form a copy of $\mathbb{C}$ inside $A$, and it is very common to then abuse notations a little and write as if we had $\mathbb{C}\subset A$. In that case, we could define $x\cdot \lambda$ by the internal multiplication of $A$, and also $\lambda^*$. But the good new is that in this situation the scalars commute with everything, so $x\cdot \lambda = \lambda\cdot x$ (with either the internal or external multiplication), and also $\lambda^* = \bar{\lambda}$.
Or, if one wants to be very precise: $x\cdot (\lambda 1_A) = (\lambda 1_A) \cdot x = \lambda\cdot x$, and $(\lambda 1_A)^* = \bar{\lambda} 1_A$.
