For the group $SU(n)$ I can write its elements in the form:
$$ U(n)= \exp(-in^a \lambda_a)$$
Considering $n^a$ to be very small I write:
$$ dU = \exp(-i dn^a \lambda_a)$$
and knowing that $dU$ is anti-Hermitian:
$$ dU = -dU^{\dagger} \Leftrightarrow \exp(-i dn^a \lambda_a) = -\exp(i dn^a \lambda_a^{\dagger})$$
I cant just apply the logarithm in both sides thanks to that negative sign, so how do I get that the generators for this group $\lambda_a = \lambda_a^{\dagger}$ are Hermitian?
Also, to get to the condition that their trace is zero I know that:
$$ det(1 +dU) = 1 + Tr(dU) = 1 \Rightarrow Tr(dU) = 0 \space\space\space\space\space\space(1)$$
which then leads to their trace being zero, but I dont understand the equalities in (1). How do I do this?
There's another way to get to hermitian-ness of the generators. If $n^a$ is very small, then I hope you don't mind if I call it $\epsilon^a$ instead. Then, we can Taylor the first exponential: $$U(\epsilon)=1-i\epsilon^a \lambda_a$$ Then use unitarity for U (the inverse is the conjugate transpose): $$\begin{align*}U(\epsilon)\,U(\epsilon)^\dagger=1&=(1-i\epsilon^a \lambda_a)(1+i\epsilon^a \lambda_a^\dagger)\\ &=1-i\epsilon^a\lambda_a+i\epsilon^a\lambda_a^\dagger\\ \end{align*}$$ Hence, $$ \lambda_a=\lambda_a^\dagger $$
For the traceless part, well, first notice that when you exponentiate a matrix, the eigenvalues all get exponentiated too. This is easy to see if you diagonalize it first.
Now consider some particular generator matrix $\lambda$. Then $U=\exp(-ik\lambda)$ is special unitary for any $k$, meaning its determinant is one. The determinant is the product of eigenvalues, so $\text{det}\,U=1=e^{-ikd_1}e^{-ikd_2}\cdots)$, where $d_i$ are the eigenvalues of $\lambda$. This means $-ik(d_1+d_2+\ldots)=0$, so $\lambda$ is traceless.