Evaluate the expression (\sqrt(-8))/(\sqrt(-9\sqrt(-2))) and write the result in the form a+bi. Real number a equals:   Real number b equals:

Let's evaluate the expression \(\frac{\sqrt{-8}}{\sqrt{-9\sqrt{-2}}}\) and express the result in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

### Step 1: Simplify the square roots involving negative numbers.
Recall that \(\sqrt{-x} = \sqrt{x} \cdot i\), where \(i\) is the imaginary unit (\(i^2 = -1\)).

So, we have:
\[
\sqrt{-8} = \sqrt{8} \cdot i
\]
\[
\sqrt{-9\sqrt{-2}} = \sqrt{9\sqrt{2} \cdot i} = \sqrt{9\sqrt{2}} \cdot \sqrt{i}
\]
Since \( \sqrt{9\sqrt{2}} = 3\sqrt{\sqrt{2}}\), and \(\sqrt{i} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\) (using the polar form of \(i\), where \(i = e^{i\frac{\pi}{2}}\)).

Thus:
\[
\sqrt{-9\sqrt{-2}} = 3\sqrt{\sqrt{2}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right) = 3\left(\frac{1}{2}\sqrt{2} + \frac{1}{2}i\sqrt{2}\right)
\]

### Step 2: Substitute back into the expression.
Now, substituting the simplified forms back into the original expression:
\[
\frac{\sqrt{-8}}{\sqrt{-9\sqrt{-2}}} = \frac{\sqrt{8} \cdot i}{3\left(\frac{1}{2}\sqrt{2} + \frac{1}{2}i\sqrt{2}\right)}
\]

### Step 3: Simplify the expression further.
\[
\frac{\sqrt{8} \cdot i}{3\left(\frac{1}{2}\sqrt{2} + \frac{1}{2}i\sqrt{2}\right)} = \frac{\sqrt{8} \cdot i}{3\left(\frac{1}{2}\sqrt{2}\right)(1 + i)}
\]
\[
= \frac{\sqrt{8} \cdot i}{\frac{3\sqrt{2}}{2}(1 + i)} = \frac{2\sqrt{2} \cdot i}{\frac{3\sqrt{2}}{2}(1 + i)} = \frac{4i}{3(1 + i)}
\]

### Step 4: Rationalize the denominator.
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator \(1 - i\):
\[
\frac{4i}{3(1 + i)} \times \frac{1 - i}{1 - i} = \frac{4i(1 - i)}{3(1 + i)(1 - i)} = \frac{4i - 4i^2}{3(1 - i^2)}
\]
Since \(i^2 = -1\), we get:
\[
= \frac{4i - 4(-1)}{3(1 - (-1))} = \frac{4i + 4}{3 \cdot 2} = \frac{4(1 + i)}{6} = \frac{2}{3} + \frac{2}{3}i
\]

### Final Answer:
\[
a = \frac{2}{3}, \quad b = \frac{2}{3}
\]

Real number \(a\) equals \(\frac{2}{3}\).

Real number \(b\) equals \(\frac{2}{3}\).

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:
1. How do you simplify expressions involving imaginary numbers?
2. What is the polar form of a complex number?
3. How do you rationalize a denominator with a complex number?
4. How is the square root of a negative number defined in complex analysis?
5. What are the properties of the imaginary unit \(i\)?
6. How do you multiply and divide complex numbers?
7. How do you convert between rectangular and polar forms of complex numbers?
8. What is the significance of the conjugate in complex number operations?

**Tip:** When working with complex numbers, it's often useful to rationalize the denominator by multiplying by the conjugate to simplify the expression.

Learn how to evaluate the complex expression \( \frac{\sqrt{-8}}{\sqrt{-9\sqrt{-2}}} \) step-by-step. Understand the use of imaginary numbers, square roots of negative numbers, and the polar form of complex numbers in this advanced high school-level problem.

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