Proof of $\sigma(zA) = z\sigma(A)$ Let $\mathfrak{U}$ be a $C^\star$-algebra with unity $I$ and let $\sigma(A)$ be the spectrum of an operator $A \in \mathfrak{U}$. 

$\sigma(A)$ is the set of all complex numbers $\lambda$ such that $A-I\lambda$ has not (two-sided) inverse.

I want to prove that, for a $z \in \mathbb{C}$, we have that  

$$\sigma(zA) = z\sigma(A)$$

Where $z\sigma(A):= \{z\lambda \in \mathbb{C}: \lambda \in \sigma(A)\}$.

My attempts: Let $x \in z\sigma(A):= \{z\lambda \in \mathbb{C}: \lambda \in \sigma(A)\}$ this implies that $x = z\lambda_x$ for some $\lambda_x \in \sigma(A)$, wich is the same as saying that $A - \lambda_x I$ has not an inverse. Suppose $z \neq 0$ then this implies that: 
$A - (x/z)I$ doesn't have an inverse iff 
$ z^{-1}(zA-xI)$ doesn't have an inverse iff 
$zA-xI$ doesn't have an inverse.
So $x \in \sigma(zA)$. Then $z\sigma(A)\subset\sigma(zA)$.
Let now $x \in \sigma(zA)$ then we have the situation before stated in such a way that if $(zA-xI)$ does not have an inverse then $x/z := \lambda \implies x = z\lambda$ where $\lambda \in \sigma(A)$ and we get our solution.

Questions: what if $z=0$? This would let $x = 0$ and something like $\sigma(zA=0)$? There is any problem in my proof? Can someone provide a rigorous (hints of a ) demonstration? This is important in proving a theorem in quantum mechanics. The second part of the proof (if correct) seems dispensable to me because the first part seems like iff cases only, is this true? 

The answer by Andre' is the same as the Rafael's proof and is only valid for the case $z \ne 0$, because the assertion
$$(A - \lambda I) \ \ \text{ is invertible} \ \ \iff z(A - \lambda I) \ \ \text{ is invertible}$$ is valid only if $z \ne 0$.
Let's pass to the case $z = 0$, which was the focus of the question.  In this case $z A = 0$, the zero element of the $C^*$ algebra, and the spectrum of the zero operator is $\{0\}$, because $0 - \lambda I = - \lambda I$ is invertible for $\lambda \ne 0$.   On the other hand, and this is the only place that some analysis is used, it is a basic theorem of functional analysis that for any $A$, $\sigma(A)$ is a non-empty compact subset of $\mathbb C$. Hence $0 \sigma(A) = \{0\}$. 
