Question

simplify the expression to ( a + bi ) form: (-sqrt{36} + sqrt{-128} - sqrt{36} + sqrt{-200})

Explanation:

Step1: Simplify square roots of negative numbers

Recall that \(\sqrt{-n}=i\sqrt{n}\) for \(n>0\). Also, simplify the real square roots.
For \(-\sqrt{36}\), \(\sqrt{36} = 6\), so \(-\sqrt{36}=- 6\).
For \(\sqrt{-128}\), \(\sqrt{-128}=i\sqrt{128}=i\sqrt{64\times2}=i\times8\sqrt{2} = 8i\sqrt{2}\).
For the second \(-\sqrt{36}\), it is also \(-6\).
For \(\sqrt{-200}\), \(\sqrt{-200}=i\sqrt{200}=i\sqrt{100\times2}=i\times10\sqrt{2}=10i\sqrt{2}\).

Step2: Combine like terms (real and imaginary)

Now, substitute these back into the original expression:

$$
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$$

Combine the real parts: \(-6-6=-12\)
Combine the imaginary parts: \(8i\sqrt{2}+10i\sqrt{2}=(8 + 10)i\sqrt{2}=18i\sqrt{2}\)

Step3: Write in \(a + bi\) form

Putting the real and imaginary parts together, we get \(-12+18\sqrt{2}i\) (or \(-12 + i18\sqrt{2}\))

Answer:

\(-12 + 18\sqrt{2}i\)