Simplifying Complex Numbers Worksheet Answer Key - Fill and Sign ... - Free Printable

Let's solve each problem on the "Simplifying Complex Numbers - Independent Practice Worksheet" step by step. The goal is to simplify expressions involving square roots of negative numbers, which involve imaginary numbers, specifically $ i $, where:

$$
i = \sqrt{-1}
$$

We'll use the rule:
$$
\sqrt{-a} = \sqrt{a} \cdot i \quad \text{(for } a > 0\text{)}
$$

Also recall:
- $ \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $

---

Problem 1: $ \sqrt{-75} $

Break down 75 into prime factors:
$$
75 = 25 \times 3 = 5^2 \times 3
$$
So:
$$
\sqrt{-75} = \sqrt{-1 \cdot 75} = \sqrt{-1} \cdot \sqrt{75} = i \cdot \sqrt{75}
$$
$$
\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}
$$
$$
\Rightarrow \sqrt{-75} = 5\sqrt{3}i
$$

✔ Answer: $ 5\sqrt{3}i $

---

Problem 2: $ i^7 $

Use powers of $ i $:
- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $
- $ i^5 = i $
- $ i^6 = -1 $
- $ i^7 = -i $

Alternatively: divide exponent by 4 and use remainder.
$$
7 \div 4 = 1 \text{ remainder } 3 \Rightarrow i^7 = i^3 = -i
$$

✔ Answer: $ -i $

---

Problem 3: $ \sqrt{-171} + \sqrt{-18} $

First, simplify each term.

#### $ \sqrt{-171} $
$$
\sqrt{-171} = \sqrt{-1 \cdot 171} = i \cdot \sqrt{171}
$$
Factor 171:
$$
171 = 9 \cdot 19 = 3^2 \cdot 19
\Rightarrow \sqrt{171} = \sqrt{9 \cdot 19} = 3\sqrt{19}
\Rightarrow \sqrt{-171} = 3\sqrt{19}i
$$

#### $ \sqrt{-18} $
$$
\sqrt{-18} = i \cdot \sqrt{18}
$$
$$
18 = 9 \cdot 2 \Rightarrow \sqrt{18} = 3\sqrt{2}
\Rightarrow \sqrt{-18} = 3\sqrt{2}i
$$

Now add:
$$
\sqrt{-171} + \sqrt{-18} = 3\sqrt{19}i + 3\sqrt{2}i = 3(\sqrt{19} + \sqrt{2})i
$$

✔ Answer: $ 3(\sqrt{19} + \sqrt{2})i $

---

Problem 4: $ \sqrt{48} - \sqrt{12} $

Both are real numbers (no negatives), so simplify radicals.

#### $ \sqrt{48} $
$$
48 = 16 \cdot 3 \Rightarrow \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}
$$

#### $ \sqrt{12} $
$$
12 = 4 \cdot 3 \Rightarrow \sqrt{12} = 2\sqrt{3}
$$

Subtract:
$$
4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}
$$

✔ Answer: $ 2\sqrt{3} $

---

Problem 5: $ \sqrt{-125} - \sqrt{25} $

#### $ \sqrt{-125} $
$$
\sqrt{-125} = i \cdot \sqrt{125}
$$
$$
125 = 25 \cdot 5 \Rightarrow \sqrt{125} = \sqrt{25 \cdot 5} = 5\sqrt{5}
\Rightarrow \sqrt{-125} = 5\sqrt{5}i
$$

#### $ \sqrt{25} = 5 $

Now subtract:
$$
\sqrt{-125} - \sqrt{25} = 5\sqrt{5}i - 5
$$

This is a complex number: $ -5 + 5\sqrt{5}i $

✔ Answer: $ -5 + 5\sqrt{5}i $

---

Problem 6: $ i^6 $

From earlier pattern:
- $ i^4 = 1 $
- $ i^5 = i $
- $ i^6 = i^2 = -1 $

Or: $ 6 \div 4 = 1 $ remainder $ 2 \Rightarrow i^6 = i^2 = -1 $

✔ Answer: $ -1 $

---

Problem 7: $ i^9 $

$ 9 \div 4 = 2 $ remainder $ 1 \Rightarrow i^9 = i^1 = i $

✔ Answer: $ i $

---

Problem 8: $ \sqrt{-64} $

$$
\sqrt{-64} = \sqrt{-1 \cdot 64} = i \cdot \sqrt{64} = i \cdot 8 = 8i
$$

✔ Answer: $ 8i $

---

Problem 9: $ \sqrt{-150} $

$$
\sqrt{-150} = i \cdot \sqrt{150}
$$
Factor 150:
$$
150 = 25 \cdot 6 = 5^2 \cdot 6
\Rightarrow \sqrt{150} = 5\sqrt{6}
\Rightarrow \sqrt{-150} = 5\sqrt{6}i
$$

✔ Answer: $ 5\sqrt{6}i $

---

Problem 10: $ \sqrt{-119} $

$$
\sqrt{-119} = i \cdot \sqrt{119}
$$
Check if 119 can be simplified:
$$
119 = 7 \cdot 17 \quad \text{(both primes)}
\Rightarrow \sqrt{119} \text{ cannot be simplified}
$$

So:
$$
\sqrt{-119} = \sqrt{119}i
$$

✔ Answer: $ \sqrt{119}i $

---

✔ Final Answers Summary:

1. $ \sqrt{-75} = 5\sqrt{3}i $
2. $ i^7 = -i $
3. $ \sqrt{-171} + \sqrt{-18} = 3(\sqrt{19} + \sqrt{2})i $
4. $ \sqrt{48} - \sqrt{12} = 2\sqrt{3} $
5. $ \sqrt{-125} - \sqrt{25} = -5 + 5\sqrt{5}i $
6. $ i^6 = -1 $
7. $ i^9 = i $
8. $ \sqrt{-64} = 8i $
9. $ \sqrt{-150} = 5\sqrt{6}i $
10. $ \sqrt{-119} = \sqrt{119}i $

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💡 Tips for solving these:
- Always factor out $ -1 $ when dealing with square roots of negative numbers.
- Simplify radicals first (factor perfect squares).
- Use the cycle of $ i $: $ i, -1, -i, 1 $, repeating every 4 powers.
- Combine like terms only (real with real, imaginary with imaginary).

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