The operator $T+i Id$ I have the operator  $$T:\mathcal{H}\rightarrow\mathcal{H}$$
the operator $T$ is linear, symmetric and compact, $\mathcal{H}$ is a Hilbert space
my problem is  

Prove that ($\,T+i\,$Id) is $\,\bf{surjective}$.   

($i$ is a imaginary number and Id is the identity operator)
I proved that $\,\,\,T+i\bf{Id}$  is injective
occupying the above, is there any way to show that $\,\,\,T+i\bf{Id}\,$ is $\,\bf{surjective}$? 
So you can give me some hint, thanks!
 A: The spectrum of $T$ is contained in the real line. Hence $-i$ is not in the spectrum This means $T+iI$ is invertible.
A: You can show that $(T-\lambda I)$ is invertible for all $\lambda\notin\mathbb{R}$. Using the fact that $\langle Tx,x\rangle$ is real, it is easy to see that
$$
         \Im\langle (T-\lambda I)x,x\rangle = -\Im\lambda \langle x,x\rangle,\;\;\; x\in \mathcal{H}
$$
Therefore,
$$
     |\Im\lambda|\|x\|^2 \le \|(T-\lambda I)x\|\|x\|, \\
      |\Im\lambda|\|x\| \le \|(T-\lambda I)x\|.
$$
So $T-\lambda I$ has trivial null space for $\lambda\notin\mathbb{R}$. 


*

*The range of $T-\lambda I$ is closed because, if $(T-\lambda I)x_n\rightarrow y$, then $(T-\lambda I)x_n$ is Cauchy, which then forces $\{x_n\}$ to be Cauchy by the last inequality. Hence, $\{x_n\}$ converges to some $x\in\mathcal{H}$ and the above implies that $(T-\lambda I)x_n\rightarrow (T-\lambda I)x=y$.

*The range of $T-\lambda I$ is $\mathcal{H}$ because $z\perp \mathcal{R}(T-\lambda I)$ implies
$$0=\langle (T-\lambda I)x,z\rangle
=\langle x,(T-\overline{\lambda}I)z\rangle \;\;\;\forall x \\
    \implies (T-\overline{\lambda}I)z=0 \\
    \implies z = 0.
$$
Therefore $T-\lambda I$ is surjective for all non-real $\lambda$ because its range is dense and its range is closed. Note: you do not need compactness of $T$ for this argument to work, only boundedness.
