Adjoint of the linear transformation over a vector space over complex numbers. Consider the vector space $V:=P_2(\mathbb{C})$ consisting of polynomials of degree at most $2$ with complex coefficients together with the following inner product $$\langle f, g\rangle=\int_{-1}^1f(t)\overline{g(t)}\, dt$$
Evaluate the adjoint of the linear transformation $T:V\to V$ by $$T(f)=if'+2f.$$
My effort:
I know that $T^*$ is a linear transformation which satisfies
$$\langle Tf, g\rangle=\langle f, T^*g\rangle.$$
Above implies that $$i\int_{-1}^1f'(t)\overline{g(t)}\, dt+2\int_{-1}^1f(t)\overline{g(t)}\, dt=\langle f, T^*g\rangle.$$
Please tell how to proceed next?

It is enough to consider the transformation $D(f)=f'$, since $T=iD+2\operatorname{id}$ yields $T^*=-iD^*+2\operatorname{id}$.
Instead of constructing an orthonormal basis for $P_2(\mathbb C)$ first, we can do the computation in our favorite basis $B=(1,t,t^2)$ using the Gram matrix of the given inner product.
The Gram matrix of $\langle -,-\rangle$ with respect to $B$ is obtained from the integrals $\int_{-1}^1 t^k\,\mathrm dt$ for $k=0,1,2,3,4$ as
$$
G_B = (\langle t^i, t^j\rangle)_{i,j=0,1,2} =
\begin{pmatrix}
2 & 0 & 2/3 \\
0 & 2/3 & 0 \\
2/3 & 0 & 2/5
\end{pmatrix}.
$$
Denoting the coordinate vector of $f\in P_2(\mathbb C)$ with respect to $B$ by $[f]_B$ this describes our inner product as
$$
\langle f,g\rangle = [f]_B^T \ G_B \ \overline{[g]_B}.
$$
The matrix of $D$ with respect to $B$ is given by
$$
[D]_B =
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 2 \\
0 & 0 & 0
\end{pmatrix}.
$$
Now by comparing the descriptions
\begin{align*}
\langle Df,g\rangle &= [f]_B^T\ [D]_B^T \ G_B \ \overline{[g]_B} \qquad\text{and} \\
\langle f,D^*g\rangle &= [f]_B^T\ G_B \ \overline{[D^*]_B}\ \overline{[g]_B},
\end{align*}
one obtains the general formula for the matrix of the adjoint in non-orthonormal bases:
$$
[D^*]_B = \overline{G_B^{-1} [D]_B^T \ G_B} =
\begin{pmatrix}
0 & -5/2 & 0 \\
3 & 0 & 1 \\
0 & 15/2 & 0
\end{pmatrix}.
$$
Hence $D^*$ is given by
$$
D^*(1) = 3t,\quad D^*(t) = -\frac 5 2+ \frac{15}2 t^2,\quad D^*(t^2) = t
$$
which translates to $T^*$ as
$$
T^*(1) = -3it+2,\quad D^*(t) = \frac 5 2i - \frac{15}2 i t^2 + 2t,\quad D^*(t^2) = -it+2t^2.
$$
