Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 10F / i / Solution

The real and imaginary parts of the operator are uniquelyThey are self-adjoint operators, and $\alpha ={\alpha }_1+i{\alpha }_2$. Conversely such a decomposition has ${\alpha }^{\ast }={\alpha }_1-i{\alpha }_2$. Expanding both products givesConsequently $\alpha$ is a normal operator exactly when its self-adjoint real and imaginary parts commute.