does $\|x\|=\sqrt{\|a\|^2+\|b\|^2}$ hold in an arbritrary $C^*$algebra?

Question

$A$ $C^*$-algebra, $x\in A$ can be written as $x=a+ib$ with $a$ and $b$ self-adjoint. Then $\|x\|=\sqrt{\|a\|^2+\|b\|^2}$ is not true in general, right? If $A=M_2(\mathbb{C})$ maybe you can find a counterexample, $\|a\|$ and $\|b\|$ is the spectral radius, I think.

Answer 1

It fails even for normal operators. Let $$ x=\begin{bmatrix}1&0\\0&i\end{bmatrix}. $$ Then $$ x=\begin{bmatrix}1&0\\0&0\end{bmatrix}+i\,\begin{bmatrix}0&0\\0&1\end{bmatrix}. $$ We have $$ \|x\|=1,\ \ \|a\|=1,\ \ \|b|=1. $$

With the same idea, it fails in every commutative C$^*$-algebra other than $\mathbb C$: in $\mathbb C^2$, take $$ x=(1,i)=(1,0)+i(0,1). $$