question\nsimplify the expression to a + bi form:\n\\(\\sqrt{16}-\\sqrt{-2}+\\sqrt{16}+\\sqrt{-200}\\)

Explanation:

Step1: Simplify square - roots of positive numbers

$\sqrt{16}=4$, so the expression becomes $4-\sqrt{-2}+4 + \sqrt{-200}$.

Step2: Rewrite square - roots of negative numbers in terms of $i$

Recall that $\sqrt{-x}=i\sqrt{x}$ for $x>0$. So, $\sqrt{-2}=i\sqrt{2}$ and $\sqrt{-200}=i\sqrt{200}=i\sqrt{100\times2}=10i\sqrt{2}$.

Step3: Combine like terms

$(4 + 4)+(-\sqrt{2}+10\sqrt{2})i=8 + 9\sqrt{2}i$.

Answer:

$8 + 9\sqrt{2}i$