For $U$ unitary and $U+iI$ hermetic, prove: $U = -iI$ Let $U$ be a unitary matrix $nxn$ such that $U+iI$ is hermetian. 
Prove: 
$$
U = -iI
$$
What i tried (And got stuck...)
If i show that $\forall u \in V$:
$$
<(U+iI)u,u> = 0
$$
Because (U+iI) is hermetic, i will be able to conclude: 
$$
(U+iI) = 0 \Rightarrow U = -iI
$$
As needed. 
So: 
$$
<(U+iI)u,u> = \\ <Uu+iu,u> = \\ <Uu,u> + i<u,u> = \\ <Uu,u>+i||u||^2 =  \\<Uu,u>+i||Tu||^2 = \\ <Uu,u> + i<Uu,Uu> 
$$
But i dont see how can i get to zero here...
Another way that i tried is using hermetic rules: 
$$
\forall u,v \in V
<(U+iI)u,u> = <u,(U+iI)u>
$$
I got to: 
$$
2i<v,u> = <v,Uu> - <Uv,u>
$$
I tried to choose $v = u$, but still i dont get too much...
Another try was to look at the norm
$$
||(U+iI)|| = <U+iI,U+iI> 
$$
I whished it will turn out zero, but i got: 
$$
<U+iI,U+iI>  = <U,U> + <I,I>
$$
Now what? 
Those are my homework, so better a hint than a solution. 
Thanks for all the answers. 
 A: Do you know the Spectral Theorem? If so, since the matrix $A:=U+iI$ is Hermitian, you know that there exist a diagonal matrix $D=diag(\lambda_1,...,\lambda_n)$ consisting of the eigenvalues for $A$ and an invertible $P$ such that $A=PDP^{-1}$. Now try to use/show that
(i) Eigenvalues of a Hermitian matrix are real and 
(ii) Eigenvalues of unitary matrices have modulus 1,
to conclude that all the $\lambda_i$'s are zero. 
A: Hint: Show that the spectral radius of $U+iI=0$. Since $U+iI$ is Hermitian this is enough prove the result. You will need the facts that spectrum of a unitary operator is contained in the unit circle and the spectrum of a Hermitian operator is contained in the real line. 
Note: the spectrum of a matrix is nothing but the set of eigen values and spectral radius is the maximum value of$|\lambda|$ over all eigen values $\lambda$. 
A: Hint: show that any element $A$ can be uniquely written as $X+iY$ with $X,Y$ Hermitian (or hermetic, however you like to call them, as long as this means $X=X^*, Y=Y^*$) (take $X=(A+A^*)/2, Y=(A-A^*)/2i)$
Write $U=(U+iI)+i(-I)$. Conclude that $U+U^*=2(U+iI)$ and therefore $U^*=U+2iI$. Multiplying by $U$, we have $U^2+2iU-I=0$. Thus $(U+iI)^2=0$.
Can you see why $U+iI=0$ from here?
