Converting an NFA to a DFA

Question

I am trying to convert this NFA to DFA:

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So I built the power automata, and this is what I got:

This should be the answer:

I don't understand where am I wrong since $$\delta_{\text{ND}}(q_1q_2,a)=\delta_{\text{D}}(q_1,a)\cup(q_2,a)=\{\epsilon\}\cup\{q_2\}=\{q_2\}$$

Who Is right? me or the book?

Answer 1

According to your picture $\delta_{ND}(q_2,a) = \{q_1, q_2\}$. Therefore, $\delta_{D}(\{q_1, q_2\},a) = \delta_{ND}(q_1,a) \cup \delta_{ND}(q_2,a) = \{q_1, q_2\}$.