= Solution
The real and imaginary parts of the operator are uniquely
$$
\alpha_1=\frac{\alpha+\alpha^*}{2},\qquad
\alpha_2=\frac{\alpha-\alpha^*}{2i}.
$$
They are <self-adjoint operators>, and $\alpha=\alpha_1+i\alpha_2$. Conversely such a decomposition has $\alpha^*=\alpha_1-i\alpha_2$. Expanding both products gives
$$
\alpha\alpha^*-\alpha^*\alpha=-2i(\alpha_1\alpha_2-\alpha_2\alpha_1).
$$
Consequently \b[$\alpha$ is a normal operator exactly when its self-adjoint real and imaginary parts commute].
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