Let $A$ be a nonunital $C^*$-algebra, $A^1=A\oplus \mathbb{C}$ as $\mathbb{C}$-vector space. Let $A^1$ be endowed with $(a+\lambda1)^*=a^*+\overline{\lambda}1$ as the involution and $(a+\lambda1)(b+\mu1)=ab+\mu a+\lambda b+\lambda\mu$ as the multiplication. We obtain a unital *-Algebra with unit $(0,1)$. A $C^*$-norm in $A^1$ is defined as follows: consider following *-homomorphism: $$L:A^1\to B(A),\; L_{(a+\lambda 1)}(b)=ab+\lambda b\in A$$for $a+\lambda 1\in A^1$ and $b\in A$. We define $$\|a+\lambda 1\|_{A^1}:=\|L_{(a+\lambda 1)}\|_{op}.$$ How to prove that $A^1$ is complete w.r.t. this norm?
My guess is to start as follows: Let $(x_n)_n\subseteq A^1$ be a cauchy sequence, write $x_n=a_n+\lambda_n1$ for $n\in\mathbb{N}$.
Can I conclude that $(a_n)_n$ is a cauchy sequence in $A$ and $(\lambda_n)_n$ a Cauchy sequence in $\mathbb{C}$?
If yes, then the completeness for $A^1$ follows, since: Is $a$ the limit of $(a_n)_n$ in $A$ and $\lambda$ the limit of $(\lambda_n)_n$ in $\mathbb{C}$, write $x:=a+\lambda1$. For every $b\in B$, it follows:
$\|(a_n-a)b+(\lambda_n -\lambda)b\|_{A}\le \|(a_n-a)\|\|b\|+|\lambda_n -\lambda|\|b\|\to 0$ for $n\to\infty$.
If not, how to prove the completeness?
Note that the map $$ L : A\to B(A) \text{ given by } a \mapsto L_a $$ is an isometry, since $\|a\| = \sup\{\|ab\| : \|b\| \leq 1\}$ by the $C^{\ast}$-identity. Hence, the image of $A$ under $L$ is a closed subspace of $B(A)$. Now note that $$A^1 = L(A) + \mathbb{C}\cdot 1_{B(A)}$$ so it is closed since it the sum of a closed subspace and a finite dimensional subspace.