Let $A$ be a bounded normal operator in a Hilbert space. Then we know there exists a continuous functional calculus defining $f(A)$ for $f\in C(\sigma(A))$ in a reasonable way. But there is also the Riesz functional calculus for functions $f$ that are holomorphic in a neighborhood of $\sigma(A)$. We have $$ f(A) = \frac 1 {2\pi i}\int_C f(z)(z-A)^{-1}\,dz, $$ where $C$ is, e.g., a circle around the spectrum of $A$.
My question: If $f$ is now only continuous on $\sigma(A)$, it makes sense to define the operator $$ T = \frac 1 {2\pi i}\int_C f(z)(z-A)^{-1}\,dz. $$ But does $T$ coincide with $f(A)$, defined through the continuous functional calculus? More specific: Do we have (in the case $f(z) = \overline z$) $$ A^* = \frac 1 {2\pi i}\int_C \overline z(z-A)^{-1}\,dz\,? $$ Thanks in advance for answers.