Let $\chi$ be the character of a $d$-dimensional complex representation $\rho$ of a finite group $G$. Then $\vert \chi (g)\vert \le d$ for all $g$ and if we have equality (for all $g$) then $\rho(g)=\lambda I$ for some root of unity ($I$ is the identity matrix).
The inequality this follows from the fact that $\chi$ is a direct sum of irreducible characters and the orders of the irreducible characters have degree dividing $d$.
May I have a hint for the second statement? I am not sure how to proceed. I know that equality implies that there will be roots of unity in the character table, but I am not sure how to show that the representation is diagonal.
It's not true that all the irreducible subrepresentations of $\rho$ must have degree dividing $d$. For instance, if $\rho_1$ is an irreducible representation of degree $1$ and $\rho_2$ is an irreducible representation of degree $2$, then $\rho=\rho_1\oplus\rho_2$ has degree $3$ but contains $\rho_2$ as a subrepresentation.
Here's a sketch of how you can show both statements instead. What you'll actually show is that if $A$ is any $d\times d$ matrix of finite order then $|\operatorname{tr}(A)|\leq d$, with equality iff $A=\lambda I$ for $\lambda$ a root of unity. To show this, note that any matrix of finite order is diagonalizable, and its eigenvalues are roots of unity. In particular, the absolute value of all the eigenvalues is $1$. Now just use the triangle inequality: the sum of $d$ numbers of absolute value $1$ has absolute value $<d$ unless they are all the same, in which case it is $d$.