[proofplan] The proof is a complex-analytic argument. Define the entire $A$-valued function $F(\zeta) := \exp(\zeta x^*) \, z \, \exp(-\zeta y^*)$. Step 1 propagates $xz = zy$ to $\exp(\eta x) z = z \exp(\eta y)$ for all $\eta \in \mathbb{C}$ by induction. Step 2 uses the normality of $x$ and $y$ together with this propagated identity to rewrite $F$ in the skew-Hermitian form $F(\zeta) = \exp(\zeta(x^* - x)) \, z \, \exp(\zeta(y - y^*))$. Step 3 bounds $\|F\|$ by $\|z\|$ on the real axis (because $\zeta(x^* - x)$ is anti-Hermitian for real $\zeta$, so the exponential is unitary in the $C^*$-algebra) and on the imaginary axis (because $i\sigma(x^* - x)$ becomes Hermitian, and $\exp(i \cdot \text{Hermitian})$ is unitary). Step 4 promotes boundedness on the real and imaginary axes to global boundedness by applying Phragmén–Lindelöf in each of the four open quadrants separately, where each quadrant is an angular sector of opening $\pi/2 < \pi$ and the exponential-type growth is below the critical Phragmén–Lindelöf threshold. Liouville then forces $F$ to be constant on $\mathbb{C}$, after passing to scalars via Hahn–Banach. Step 5 differentiates $F$ at $0$ to extract $x^* z = z y^*$. [/proofplan] [step:Define the entire $A$-valued function $F$] The element $\zeta x^* \in A$ for $\zeta \in \mathbb{C}$ has the convergent exponential series \begin{align*} \exp(\zeta x^*) := \sum_{n=0}^\infty \frac{(\zeta x^*)^n}{n!} \in A, \end{align*} absolutely convergent because $\sum_{n=0}^\infty \|\zeta x^*\|^n / n! = \exp(|\zeta| \|x^*\|) < \infty$, and the partial sums form a Cauchy sequence in the Banach algebra $A$. Similarly $\exp(-\zeta y^*) \in A$ is well-defined for every $\zeta \in \mathbb{C}$. Both $\zeta \mapsto \exp(\zeta x^*)$ and $\zeta \mapsto \exp(-\zeta y^*)$ are entire $A$-valued power series — series with $A$-valued coefficients converging on $|\zeta| < \infty$. Define \begin{align*} F: \mathbb{C} &\to A \\ \zeta &\mapsto \exp(\zeta x^*) \, z \, \exp(-\zeta y^*). \end{align*} Multiplication in a Banach algebra is jointly continuous and bilinear, so the product of two entire $A$-valued power series with the constant element $z$ inserted between them is again an entire $A$-valued power series. Hence $F$ is entire. [/step] [step:Propagate the identity $xz = zy$ to $\exp(\eta x) \, z = z \, \exp(\eta y)$ for every $\eta \in \mathbb{C}$] We claim $x^n z = z y^n$ for all $n \ge 0$ by induction on $n$. *Base case $n = 0$.* $x^0 z = 1 \cdot z = z = z \cdot 1 = z y^0$, using that $A$ is unital so $x^0 = y^0 = 1$. *Inductive step.* Assume $x^n z = z y^n$. Using the hypothesis $xz = zy$ and the inductive hypothesis, \begin{align*} x^{n+1} z = x (x^n z) = x (z y^n) = (xz) y^n = (zy) y^n = z y^{n+1}. \end{align*} By induction, $x^n z = z y^n$ for all $n \ge 0$. Multiplying by $\eta^n / n!$ for any $\eta \in \mathbb{C}$ and summing over $n$: \begin{align*} \exp(\eta x) \cdot z = \left(\sum_{n=0}^\infty \frac{\eta^n x^n}{n!}\right) z = \sum_{n=0}^\infty \frac{\eta^n}{n!} (x^n z) = \sum_{n=0}^\infty \frac{\eta^n}{n!} (z y^n) = z \cdot \exp(\eta y). \end{align*} The interchange of $z$ with the infinite sum on either side is justified by continuity of left- and right-multiplication by $z$ in $A$ (these are bounded linear maps with operator norm at most $\|z\|$, hence commute with norm-convergent limits). Hence \begin{align*} \exp(\eta x) \, z = z \, \exp(\eta y) \quad \text{for all } \eta \in \mathbb{C}. \tag{$\dagger$} \end{align*} [/step] [step:Rewrite $F(\zeta) = \exp(\zeta(x^* - x)) \, z \, \exp(\zeta(y - y^*))$ using normality and $(\dagger)$] *Lemma (commuting exponentials add).* If $a, b \in A$ commute ($ab = ba$), then $\exp(a + b) = \exp(a) \exp(b) = \exp(b) \exp(a)$. The binomial theorem applied to commuting elements gives $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$; summing against $1/n!$ and reorganising by absolute convergence (Cauchy product) yields the claim. *Decompose $\exp(\zeta x^*)$.* Since $x$ is normal, $x x^* = x^* x$, so $-\zeta x$ and $\zeta x^*$ commute for every $\zeta \in \mathbb{C}$. Applying the lemma, \begin{align*} \exp(\zeta(x^* - x)) = \exp(-\zeta x + \zeta x^*) = \exp(-\zeta x) \, \exp(\zeta x^*) = \exp(\zeta x^*) \, \exp(-\zeta x). \end{align*} Multiplying on the right by $\exp(\zeta x)$ — valid since $\zeta x$ and $-\zeta x$ commute, so $\exp(-\zeta x) \exp(\zeta x) = \exp(0) = 1$ — yields \begin{align*} \exp(\zeta(x^* - x)) \, \exp(\zeta x) = \exp(\zeta x^*), \end{align*} or equivalently, \begin{align*} \exp(\zeta x^*) = \exp(\zeta(x^* - x)) \, \exp(\zeta x). \tag{$\dagger\dagger$} \end{align*} *Decompose $\exp(-\zeta y^*)$.* Similarly, normality of $y$ gives $y y^* = y^* y$, so $-\zeta y^*$ and $\zeta y$ commute. By the lemma, \begin{align*} \exp(\zeta(y - y^*)) = \exp(\zeta y - \zeta y^*) = \exp(\zeta y) \, \exp(-\zeta y^*) = \exp(-\zeta y^*) \, \exp(\zeta y). \end{align*} Multiplying on the left by $\exp(-\zeta y)$ — valid since $\zeta y$ and $-\zeta y$ commute, so $\exp(-\zeta y) \exp(\zeta y) = 1$ — yields \begin{align*} \exp(-\zeta y^*) = \exp(-\zeta y) \, \exp(\zeta(y - y^*)). \tag{$\ddagger$} \end{align*} *Substitute $(\dagger\dagger)$ and $(\ddagger)$ into $F(\zeta)$.* Plugging in, \begin{align*} F(\zeta) &= \exp(\zeta x^*) \, z \, \exp(-\zeta y^*) \\ &= \bigl[\exp(\zeta(x^* - x)) \, \exp(\zeta x)\bigr] \, z \, \bigl[\exp(-\zeta y) \, \exp(\zeta(y - y^*))\bigr] \\ &= \exp(\zeta(x^* - x)) \, \bigl[\exp(\zeta x) \, z \, \exp(-\zeta y)\bigr] \, \exp(\zeta(y - y^*)). \end{align*} By $(\dagger)$ from Step 2, $\exp(\zeta x) \, z = z \, \exp(\zeta y)$, so the bracketed factor equals \begin{align*} \exp(\zeta x) \, z \, \exp(-\zeta y) = z \, \exp(\zeta y) \, \exp(-\zeta y) = z \cdot 1 = z, \end{align*} using that $\zeta y$ and $-\zeta y$ commute, hence $\exp(\zeta y) \exp(-\zeta y) = \exp(0) = 1$. We conclude \begin{align*} F(\zeta) = \exp(\zeta(x^* - x)) \, z \, \exp(\zeta(y - y^*)) \quad \text{for all } \zeta \in \mathbb{C}. \tag{$\star$} \end{align*} Set $u := x^* - x$ and $v := y - y^*$. Both are anti-Hermitian: $u^* = (x^*)^* - x^* = x - x^* = -u$, and $v^* = y^* - (y^*)^* = y^* - y = -v$. [/step] [step:Bound $\|F(\zeta)\|$ on the real and imaginary axes via $C^*$-unitarity] *Norm-one principle.* For any anti-Hermitian element $w \in A$ (that is, $w^* = -w$) and any $\sigma \in \mathbb{R}$, the element $\sigma w$ is again anti-Hermitian, and $\exp(\sigma w)$ is unitary: \begin{align*} \exp(\sigma w)^* \, \exp(\sigma w) = \exp((\sigma w)^*) \, \exp(\sigma w) = \exp(-\sigma w) \, \exp(\sigma w) = \exp(0) = 1, \end{align*} using that the involution $* : A \to A$ is continuous and conjugate-linear, hence applies term-by-term to the exponential series and conjugates each $(\sigma w)^n$ to $((\sigma w)^*)^n = (-\sigma w)^n$, giving $\exp(\sigma w)^* = \exp(-\sigma w)$. The middle step uses that $\sigma w$ and $-\sigma w$ commute (both are scalar multiples of $w$). Similarly $\exp(\sigma w) \exp(\sigma w)^* = 1$. In a $C^*$-algebra, a unitary element has norm $1$ via the $C^*$-identity $\|c^* c\| = \|c\|^2$: \begin{align*} \|\exp(\sigma w)\|^2 = \|\exp(\sigma w)^* \, \exp(\sigma w)\| = \|1\| = 1, \quad \text{so } \|\exp(\sigma w)\| = 1. \end{align*} Likewise, for any Hermitian element $h \in A$ ($h^* = h$) and any $\tau \in \mathbb{R}$, the element $i\tau h$ is anti-Hermitian (since $(i\tau h)^* = -i\tau h^* = -i\tau h$), so $\|\exp(i\tau h)\| = 1$ by the same argument. *Boundedness on the real axis.* By $(\star)$ with $\zeta = \sigma \in \mathbb{R}$, \begin{align*} F(\sigma) = \exp(\sigma u) \, z \, \exp(\sigma v), \end{align*} where $u = x^* - x$ and $v = y - y^*$ are anti-Hermitian. The norm-one principle gives $\|\exp(\sigma u)\| = \|\exp(\sigma v)\| = 1$, so by submultiplicativity, \begin{align*} \|F(\sigma)\| \le \|\exp(\sigma u)\| \, \|z\| \, \|\exp(\sigma v)\| = \|z\| \quad \text{for all } \sigma \in \mathbb{R}. \tag{R} \end{align*} *Boundedness on the imaginary axis.* Set $\zeta = i\tau$ with $\tau \in \mathbb{R}$. Then $\zeta u = i\tau u$, and we observe that $i u = i(x^* - x)$ is Hermitian: \begin{align*} (i u)^* = -i \, u^* = -i \cdot (-u) = i u, \end{align*} using that $u$ is anti-Hermitian. So $i u$ is Hermitian. The same applies to $i v = i(y - y^*)$: $(iv)^* = -iv^* = -i(-v) = iv$, so $iv$ is Hermitian. Therefore $\zeta u = i\tau u = \tau (i u)$ with $i u$ Hermitian and $\tau \in \mathbb{R}$, and likewise $\zeta v = \tau (i v)$. By the norm-one principle applied with the Hermitian elements $iu, iv$: \begin{align*} \|\exp(\zeta u)\| = \|\exp(\tau (iu))\| = 1, \qquad \|\exp(\zeta v)\| = \|\exp(\tau (iv))\| = 1. \end{align*} Hence by $(\star)$ and submultiplicativity, \begin{align*} \|F(i\tau)\| \le \|\exp(i\tau u)\| \, \|z\| \, \|\exp(i\tau v)\| = \|z\| \quad \text{for all } \tau \in \mathbb{R}. \tag{I} \end{align*} *Global growth bound.* For any $c \in A$ and any $\zeta \in \mathbb{C}$, \begin{align*} \|\exp(\zeta c)\| \le \sum_{n=0}^\infty \frac{|\zeta|^n \|c\|^n}{n!} = \exp(|\zeta| \|c\|). \end{align*} Applying this to $(\star)$, \begin{align*} \|F(\zeta)\| \le \exp(|\zeta| \|u\|) \cdot \|z\| \cdot \exp(|\zeta| \|v\|) = \|z\| \, \exp(M|\zeta|), \qquad M := \|u\| + \|v\|. \tag{G} \end{align*} [/step] [step:Apply Phragmén–Lindelöf in each quadrant to obtain $F \equiv z$] We work via scalar Liouville: fix an arbitrary continuous linear functional $\varphi \in A^*$ and define \begin{align*} g_\varphi: \mathbb{C} &\to \mathbb{C} \\ \zeta &\mapsto \varphi(F(\zeta)). \end{align*} The composition of an entire $A$-valued function $F$ with a continuous $\mathbb{C}$-linear functional $\varphi$ is entire (the partial sums of the power series of $F$ are mapped by $\varphi$ to partial sums of a $\mathbb{C}$-valued power series with the same domain of convergence). The bounds (R), (I), (G) give \begin{align*} |g_\varphi(\sigma)| &\le \|\varphi\| \, \|z\| =: K \quad \text{for all } \sigma \in \mathbb{R}, \\ |g_\varphi(i\tau)| &\le K \quad \text{for all } \tau \in \mathbb{R}, \\ |g_\varphi(\zeta)| &\le K \, \exp(M|\zeta|) \quad \text{for all } \zeta \in \mathbb{C}. \end{align*} *Phragmén–Lindelöf in an angular sector.* The classical [Phragmén–Lindelöf principle](/theorems/???) for an angular sector (see, e.g., Rudin, *Real and Complex Analysis*, Theorem 12.8) states: if $S$ is the open sector $\{\zeta = re^{i\theta} : r > 0,\, \alpha < \theta < \beta\}$ with opening angle $\beta - \alpha < \pi$, and $G : \overline{S} \to \mathbb{C}$ is continuous on the closure, holomorphic in $S$, satisfies $|G(\zeta)| \le K$ on the two boundary rays $\arg \zeta = \alpha$ and $\arg \zeta = \beta$, and obeys a growth bound $|G(\zeta)| \le K_1 \exp(K_2 |\zeta|^p)$ throughout $S$ for some $p < \pi/(\beta - \alpha)$, then $|G(\zeta)| \le K$ throughout $\overline{S}$. *Apply to each open quadrant.* Each of the four open coordinate quadrants \begin{align*} Q_1 &= \{\zeta : 0 < \arg \zeta < \pi/2\}, & Q_2 &= \{\zeta : \pi/2 < \arg \zeta < \pi\}, \\ Q_3 &= \{\zeta : \pi < \arg \zeta < 3\pi/2\}, & Q_4 &= \{\zeta : 3\pi/2 < \arg \zeta < 2\pi\} \end{align*} is an open sector of opening $\beta - \alpha = \pi/2$. The Phragmén–Lindelöf threshold $p < \pi/(\beta - \alpha) = \pi/(\pi/2) = 2$ admits any growth of order strictly below $2$. The growth bound (G) gives $|g_\varphi(\zeta)| \le K \exp(M|\zeta|^1)$, which is of order $1 < 2$ — strictly below the threshold. Each quadrant $Q_j$ has its two boundary rays contained in $\mathbb{R} \cup i\mathbb{R}$, and on these axes $|g_\varphi| \le K$ by (R) and (I). The function $g_\varphi$ is continuous on $\overline{Q_j}$ (being entire) and holomorphic on $Q_j$. Phragmén–Lindelöf in the sector $Q_j$ applies and yields $|g_\varphi(\zeta)| \le K$ throughout $\overline{Q_j}$. Taking the union over $j = 1, 2, 3, 4$, \begin{align*} |g_\varphi(\zeta)| \le K = \|\varphi\| \, \|z\| \quad \text{for all } \zeta \in \mathbb{C}. \end{align*} *Liouville's theorem.* By [Liouville's theorem](/theorems/???), a bounded entire $\mathbb{C}$-valued function is constant. Hence $g_\varphi$ is constant, and \begin{align*} \varphi(F(\zeta)) = g_\varphi(\zeta) = g_\varphi(0) = \varphi(F(0)) = \varphi(z) \quad \text{for all } \zeta \in \mathbb{C}. \end{align*} *Pass back to $A$ via Hahn–Banach.* The identity $\varphi(F(\zeta) - z) = 0$ holds for every $\varphi \in A^*$. The continuous linear functionals on the Banach space $A$ separate points (a corollary of the [Hahn–Banach theorem](/theorems/???)): for every nonzero $w \in A$ there exists $\varphi \in A^*$ with $\varphi(w) \neq 0$. Since $F(\zeta) - z$ is annihilated by every $\varphi \in A^*$, we conclude $F(\zeta) - z = 0$, i.e., \begin{align*} F(\zeta) = z \quad \text{for all } \zeta \in \mathbb{C}. \end{align*} [guided] The strategy of Step 4 is to combine (R), (I), and (G) to force $F$ to be globally bounded, then invoke Liouville. The mechanism is Phragmén–Lindelöf in an angular sector. We now expand the argument. *Why two perpendicular lines of boundedness?* The growth bound (G) alone — $\|F(\zeta)\| \le \|z\| \exp(M|\zeta|)$ — does not suffice to force constancy: the entire function $\zeta \mapsto e^{M\zeta}$ has exactly this type of growth and is far from constant. A single line of boundedness is also not enough: $\zeta \mapsto e^{M\zeta}$ is bounded on $i\mathbb{R}$ (since $|e^{i M \tau}| = 1$ for $\tau \in \mathbb{R}$) yet still not constant. What forces constancy is boundedness on enough boundary rays to box in the exponential type from every direction, combined with a growth bound that is strictly subcritical for the angular openings between those rays. Boundedness on the real axis (R) and the imaginary axis (I) gives four boundary rays — the positive and negative real half-axes and the positive and negative imaginary half-axes. These rays cut $\mathbb{C}$ into four open quadrants $Q_1, Q_2, Q_3, Q_4$, each of opening angle $\pi/2$. In each quadrant, the boundary consists of two of these rays, and on each ray we have $|g_\varphi| \le K$. *Why does the threshold work?* Phragmén–Lindelöf in an angular sector of opening $\beta - \alpha$ tolerates growth of order strictly less than $\pi/(\beta - \alpha)$. For a quadrant ($\beta - \alpha = \pi/2$) the tolerated order is strictly less than $2$; our growth (G) is of order $1$, comfortably below. For a half-plane ($\beta - \alpha = \pi$), the threshold drops to strictly less than $1$, and exponential-type-$1$ growth like $e^{M\zeta}$ is *exactly at* the borderline — Phragmén–Lindelöf in a half-plane fails to conclude boundedness from boundedness on the boundary line, which is why a single line of boundedness combined with growth $\exp(M|\zeta|)$ is insufficient. The presence of two perpendicular lines, halving the sector openings to $\pi/2$, drops us to the strict-inequality regime where the principle applies. *The Phragmén–Lindelöf statement, applied to $Q_1$.* The classical principle in a sector of opening less than $\pi$ states: a function continuous on the closure of the open sector $S$, holomorphic in $S$, bounded on the two boundary rays, and obeying a growth bound below the critical exponential order, is bounded on $\overline{S}$ by the boundary bound. We verify each hypothesis for $g_\varphi$ on $\overline{Q_1}$: - **Holomorphy in $Q_1$:** $g_\varphi$ is entire on $\mathbb{C}$, hence holomorphic on the open quadrant $Q_1$. - **Continuity on $\overline{Q_1}$:** $g_\varphi$ is entire, hence continuous on all of $\mathbb{C}$, in particular on the closed quadrant $\overline{Q_1}$. - **Boundary boundedness:** The boundary $\partial Q_1$ consists of the closed rays $[0, \infty)$ and $i[0, \infty)$. On $[0, \infty) \subset \mathbb{R}$, bound (R) gives $|g_\varphi(\sigma)| \le K$. On $i[0, \infty) \subset i\mathbb{R}$, bound (I) gives $|g_\varphi(i\tau)| \le K$. So $|g_\varphi| \le K$ on $\partial Q_1$. - **Subcritical growth:** The opening angle is $\pi/2$, so the Phragmén–Lindelöf threshold for the order is $\pi/(\pi/2) = 2$. From (G), $|g_\varphi(\zeta)| \le K \exp(M|\zeta|)$, which is of order $1$ (the function is dominated by $\exp(M r)$ on $|\zeta| = r$, and $\log\log|g_\varphi(re^{i\theta})| / \log r \to 1$ at most). Since $1 < 2$, the growth is strictly subcritical. All hypotheses verified, Phragmén–Lindelöf in the sector $Q_1$ yields $|g_\varphi(\zeta)| \le K$ for $\zeta \in \overline{Q_1}$. The argument is identical, mutatis mutandis, in each of $Q_2, Q_3, Q_4$. The boundary of each $Q_j$ is a pair of half-axes from $\mathbb{R} \cup i\mathbb{R}$, and the bounds (R), (I) handle each. Combining, $|g_\varphi(\zeta)| \le K$ on $\bigcup_j \overline{Q_j} = \mathbb{C}$. *Concluding via Liouville and Hahn–Banach.* The bounded entire function $g_\varphi$ is constant by Liouville's theorem (any bounded entire $\mathbb{C}$-valued function is constant). Hence $g_\varphi(\zeta) = g_\varphi(0) = \varphi(F(0)) = \varphi(z)$ for every $\zeta \in \mathbb{C}$, i.e., $\varphi(F(\zeta) - z) = 0$ for every $\varphi \in A^*$. To pass from "annihilated by every continuous functional" to "equal to zero" we invoke the separating-functionals corollary of Hahn–Banach: for every nonzero element $w$ of a Banach space $X$ there exists $\varphi \in X^*$ with $\varphi(w) = \|w\| \neq 0$ (extend the unit functional on $\mathrm{span}(w)$ to all of $X$ by Hahn–Banach). Hence if $\varphi(w) = 0$ for every $\varphi \in A^*$ then $w = 0$. Applying this to $w = F(\zeta) - z$ gives $F(\zeta) = z$ for every $\zeta \in \mathbb{C}$. *Why two lines, not one — restated.* The naive attempt "establish boundedness on $i\mathbb{R}$ and apply Phragmén–Lindelöf in a half-plane" fails because the half-plane has opening $\pi$ and the critical order drops to $1$, exactly matching our growth — at the borderline, the principle requires a strictly slower growth or an extra log factor, and the function $e^{M\zeta}$ shows that the borderline is achievable with non-constant entire functions. Halving the opening to $\pi/2$ via the second line of boundedness moves us into the strict-inequality regime $1 < 2$, where the principle applies and the conclusion of constancy follows. [/guided] [/step] [step:Differentiate $F$ at zero to extract $x^* z = z y^*$] The function $F: \mathbb{C} \to A$ is entire and constant by Step 4. Hence $F'(\zeta) = 0$ for all $\zeta \in \mathbb{C}$. We compute $F'(0)$ from the original definition $F(\zeta) = U(\zeta) \, z \, V(\zeta)$ with $U(\zeta) := \exp(\zeta x^*)$ and $V(\zeta) := \exp(-\zeta y^*)$. The derivative of $\exp(\zeta c)$ in a Banach algebra is $c \exp(\zeta c) = \exp(\zeta c) c$ (these agree because $c$ commutes with every term of the power series in $c$). Hence \begin{align*} U'(\zeta) = x^* \exp(\zeta x^*), \qquad V'(\zeta) = -y^* \exp(-\zeta y^*). \end{align*} The product rule for $A$-valued functions of one complex variable (which holds because multiplication is bilinear and continuous on bounded sets) gives \begin{align*} F'(\zeta) = U'(\zeta) \, z \, V(\zeta) + U(\zeta) \, z \, V'(\zeta) = x^* \exp(\zeta x^*) \, z \, \exp(-\zeta y^*) - \exp(\zeta x^*) \, z \, y^* \exp(-\zeta y^*), \end{align*} where in the second term we used that $y^*$ commutes with $\exp(-\zeta y^*)$ (both are functions of $y^*$ alone), so $z \, V'(\zeta) = -z \, y^* \exp(-\zeta y^*)$. Evaluate at $\zeta = 0$, where $\exp(0) = 1$: \begin{align*} 0 = F'(0) = x^* \cdot 1 \cdot z \cdot 1 - 1 \cdot z \, y^* \cdot 1 = x^* z - z y^*. \end{align*} Hence $x^* z = z y^*$, completing the proof of the implication $xz = zy \implies x^* z = z y^*$. *The "in particular" specialisation.* Setting $y = x$ in the implication: for every $z \in A$, if $xz = zx$ then $x^* z = z x^*$. The converse follows by applying the same implication to the pair $(x^*, x^*, z)$: the element $x^*$ is normal because $(x^*)^* x^* = x x^* = x^* x = x^* (x^*)^*$, so the implication says $x^* z = z x^* \implies (x^*)^* z = z (x^*)^*$, i.e., $xz = zx$. Hence the equivalence $xz = zx \iff x^* z = z x^*$ holds for every $z \in A$. [/step]