To solve the questions in the highlighted section (f, g, h), let’s break them down step by step:

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(f) $\frac{1}{i} + i$

1. Recall that $\frac{1}{i}$ can be simplified by multiplying numerator and denominator by $-i$ (the conjugate of $i$): $\frac{1}{i} = \frac{1 \cdot (-i)}{i \cdot (-i)} = -i$

2. Adding $-i + i$: $-i + i = 0$

Result for (f): $\boxed{0}$

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(g) $\frac{26}{(2 - 3i)(2 + 3i)}$

1. Simplify the denominator using the formula for the product of conjugates: $(2 - 3i)(2 + 3i) = 2^2 - (3i)^2 = 4 - (-9) = 4 + 9 = 13$

2. The fraction becomes: $\frac{26}{13} = 2$

Result for (g): $\boxed{2}$

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(h) $\sqrt{\frac{-7 + i}{2} - \frac{6 - i}{5 - 7i}}$

1. Simplify each term in the fraction:

   • First term: $\frac{-7 + i}{2}$ remains as is for now.
   • Second term: Simplify $\frac{6 - i}{5 - 7i}$ by multiplying numerator and denominator by the conjugate $5 + 7i$: $(6 - i)(5 + 7i) = 30 + 42i - 5i - 7i^2 = 30 + 37i + 7 = 37 + 37i$ $(5 - 7i)(5 + 7i) = 5^2 - (7i)^2 = 25 - (-49) = 25 + 49 = 74$ $\frac{6 - i}{5 - 7i} = \frac{37 + 37i}{74} = \frac{37}{74} + \frac{37i}{74} = \frac{1}{2} + \frac{i}{2}$

2. Substitute back into the expression: $\frac{-7 + i}{2} - \left(\frac{1}{2} + \frac{i}{2}\right) = \frac{-7 + i - 1 - i}{2} = \frac{-8}{2} = -4$

3. Taking the square root: $\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i$

Result for (h): $\boxed{2i}$

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Final Results:

• (f): $0$
• (g): $2$
• (h): $2i$

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Would you like a deeper explanation for any of these steps?
Here are 5 related questions you can explore:

1. How do you simplify expressions involving imaginary units ($i$)?
2. What are conjugates of complex numbers, and how are they useful?
3. Can you explain the formula for the product of conjugates in detail?
4. How do you rationalize denominators with complex numbers?
5. What is the process for handling square roots of negative numbers?

Tip: Always use conjugates to simplify complex-number divisions.