Consider the Schur function for irrep $(1)$ of $\mathrm{U}(3)$: $s_{(1)}(z_1,z_2,z_3)=z_1+z_2+z_3$. What is the Schur function for the irrep conjugate to this, i.e. the irrep $(1)^*$ where all the entries are conjugated?
The irrep $(1)^*$ has the same dimension as $(1)$ but its weights will be the inverses of those in $(1)$, i.e. the weights will be $1/z_k$. It is cannot be equivalent to $(1)$ since any similarity transformation does not change the eigenvalues. Thus, $s_{(1)^*}(z_1,z_2,z_3)$ should be something like $$ s_{(1)^*}(z_1,z_2,z_3)=\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\, , \tag{A} $$ but there is no partition that will produce a Schur function with negative powers of the arguments. In fact, $$ s_{(1)^*}(z_1,z_2,z_3)=\frac{s_{(1,1)}(z_1,z_2,z_3)}{z_1 z_2 z_3}\, , $$ where $z_1 z_2 z_3$ is the determinant of the irrep $(1)$.
How does one then decompose and interpret $(1)^*\otimes (1)$ as the resulting product $s_{(1)^*}(z_1,z_2,z_3)s_{(1)}(z_1,z_2,z_3)$ will be the expression $$ \frac{z_1}{z_2}+\frac{z_1}{z_3}+\frac{z_3}{z_2}+\frac{z_2}{z_3}+\frac{z_2}{z_1}+\frac{z_3}{z_1}+3, \tag{B} $$ which contains negative powers of the $z_k$’s? It's not that hard to find the decomposition (which should contain the identify representation) using other methods but I’m more interested in understanding how to properly deal (in general) with inverse powers of the $z_k$’s in Schur functions.