Question

question
simplify the expression to (a + bi) form:
(sqrt{9} + sqrt{-162} + sqrt{16} + sqrt{-128})

Explanation:

Step1: Simplify real square roots

Simplify \(\sqrt{9}\) and \(\sqrt{16}\). We know that \(\sqrt{9} = 3\) and \(\sqrt{16}=4\).

Step2: Simplify imaginary square roots

For \(\sqrt{-162}\) and \(\sqrt{-128}\), use the property \(\sqrt{-x}=\sqrt{x}i\) (where \(x>0\)).
First, \(\sqrt{-162}=\sqrt{81\times2}i = 9\sqrt{2}i\) (since \(\sqrt{81}=9\)).
Second, \(\sqrt{-128}=\sqrt{64\times2}i = 8\sqrt{2}i\) (since \(\sqrt{64}=8\)).

Step3: Combine real and imaginary parts

Combine the real parts: \(3 + 4=7\).
Combine the imaginary parts: \(9\sqrt{2}i+8\sqrt{2}i=(9 + 8)\sqrt{2}i = 17\sqrt{2}i\).
Then the expression becomes \(7+17\sqrt{2}i\).

Answer:

\(7 + 17\sqrt{2}i\)