Practice worksheet for dividing complex numbers, featuring 20 algebraic problems.

Let’s solve each problem one by one. We’re dividing complex numbers — that means we have expressions like (a + bi) ÷ (c + di). To divide them, we multiply the top and bottom by the conjugate of the bottom number. The conjugate of (c + di) is (c - di). This gets rid of the imaginary part in the denominator.

We’ll go step by step for each problem.

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Problem 1: 8 / (2i)

This is simple. Just write it as a fraction and simplify.

8 / (2i) = 4 / i

Now, to remove i from the denominator, multiply numerator and denominator by i:

(4 × i) / (i × i) = 4i / (-1) = -4i

✔ Final Answer: -4i

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Problem 2: 3 / (-i)

Same idea.

3 / (-i) = -3 / i

Multiply numerator and denominator by i:

(-3 × i) / (i × i) = -3i / (-1) = 3i

✔ Final Answer: 3i

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Problem 3: (6 - i) / 5i

Write as fraction: (6 - i)/(5i)

Multiply numerator and denominator by i (since denominator is purely imaginary):

Numerator: (6 - i) × i = 6i - i² = 6i - (-1) = 6i + 1

Denominator: 5i × i = 5i² = 5(-1) = -5

So we get: (1 + 6i) / (-5) = -1/5 - (6/5)i

✔ Final Answer: -1/5 - (6/5)i

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Problem 4: 6 / (7 - 2i)

Multiply numerator and denominator by conjugate of denominator: (7 + 2i)

Numerator: 6 × (7 + 2i) = 42 + 12i

Denominator: (7 - 2i)(7 + 2i) = 7² - (2i)² = 49 - 4i² = 49 - 4(-1) = 49 + 4 = 53

So: (42 + 12i)/53 = 42/53 + (12/53)i

✔ Final Answer: 42/53 + (12/53)i

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Problem 5: (9 + i) / (2 - 5i)

Multiply numerator and denominator by (2 + 5i)

Numerator: (9 + i)(2 + 5i) = 9×2 + 9×5i + i×2 + i×5i = 18 + 45i + 2i + 5i²
= 18 + 47i + 5(-1) = 18 + 47i - 5 = 13 + 47i

Denominator: (2 - 5i)(2 + 5i) = 2² - (5i)² = 4 - 25i² = 4 - 25(-1) = 4 + 25 = 29

So: (13 + 47i)/29 = 13/29 + (47/29)i

✔ Final Answer: 13/29 + (47/29)i

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Problem 6: (5 - 4i) / (-3i)

Multiply numerator and denominator by i (to eliminate i from denominator)

Numerator: (5 - 4i) × i = 5i - 4i² = 5i - 4(-1) = 5i + 4

Denominator: -3i × i = -3i² = -3(-1) = 3

So: (4 + 5i)/3 = 4/3 + (5/3)i

✔ Final Answer: 4/3 + (5/3)i

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Problem 7: (1 - 7i) / (1 - 2i)

Multiply numerator and denominator by (1 + 2i)

Numerator: (1 - 7i)(1 + 2i) = 1×1 + 1×2i -7i×1 -7i×2i = 1 + 2i -7i -14i²
= 1 -5i -14(-1) = 1 -5i +14 = 15 -5i

Denominator: (1 - 2i)(1 + 2i) = 1² - (2i)² = 1 - 4i² = 1 - 4(-1) = 1 + 4 = 5

So: (15 - 5i)/5 = 3 - i

✔ Final Answer: 3 - i

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Problem 8: (2 - 3i) / (3 - 3i)

First, factor out 3 from denominator? Or just multiply by conjugate (3 + 3i)

But notice: denominator is 3(1 - i), so maybe simplify first? Let’s do conjugate method.

Multiply numerator and denominator by (3 + 3i)

Numerator: (2 - 3i)(3 + 3i) = 2×3 + 2×3i -3i×3 -3i×3i = 6 + 6i -9i -9i²
= 6 -3i -9(-1) = 6 -3i +9 = 15 -3i

Denominator: (3 - 3i)(3 + 3i) = 3² - (3i)² = 9 - 9i² = 9 - 9(-1) = 9 + 9 = 18

So: (15 - 3i)/18 = 15/18 - 3i/18 = 5/6 - (1/6)i

✔ Final Answer: 5/6 - (1/6)i

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Problem 9: (-4 - 2i) / (-2 + 5i)

Multiply numerator and denominator by conjugate of denominator: (-2 - 5i)

Wait — actually, conjugate of (-2 + 5i) is (-2 - 5i)? Yes.

But let’s make it easier: factor out -1 from numerator and denominator?

Numerator: -4 -2i = -2(2 + i)

Denominator: -2 + 5i → can’t factor easily. Better to proceed with conjugate.

Multiply numerator and denominator by (-2 - 5i)

Numerator: (-4 -2i)(-2 -5i) = (-4)(-2) + (-4)(-5i) + (-2i)(-2) + (-2i)(-5i)
= 8 + 20i + 4i + 10i² = 8 + 24i + 10(-1) = 8 + 24i -10 = -2 + 24i

Denominator: (-2 + 5i)(-2 -5i) = (-2)² - (5i)² = 4 - 25i² = 4 - 25(-1) = 4 + 25 = 29

So: (-2 + 24i)/29 = -2/29 + (24/29)i

✔ Final Answer: -2/29 + (24/29)i

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Problem 10: (2 + 7i) / (4 - 2i)

Multiply numerator and denominator by (4 + 2i)

Numerator: (2 + 7i)(4 + 2i) = 2×4 + 2×2i + 7i×4 + 7i×2i = 8 + 4i + 28i + 14i²
= 8 + 32i + 14(-1) = 8 + 32i -14 = -6 + 32i

Denominator: (4 - 2i)(4 + 2i) = 16 - (2i)² = 16 - 4i² = 16 - 4(-1) = 16 + 4 = 20

So: (-6 + 32i)/20 = -6/20 + 32i/20 = -3/10 + (8/5)i

Wait — 32/20 simplifies to 8/5? 32÷4=8, 20÷4=5 → yes.

But let’s write both fractions with same denominator or leave as is.

Actually: -6/20 = -3/10, 32/20 = 8/5 → but 8/5 is 1.6, which is fine.

Alternatively, write as: (-3/10) + (8/5)i

But 8/5 = 16/10, so if we want common denominator: (-3/10) + (16/10)i → but not necessary.

✔ Final Answer: -3/10 + (8/5)i

Wait — double-check calculation:

(2+7i)(4+2i) = 2*4=8, 2*2i=4i, 7i*4=28i, 7i*2i=14i² → total: 8 + 32i -14 = -6 + 32i → correct.

Denominator: 16 - (2i)^2 = 16 - (-4) = 20 → correct.

-6/20 = -3/10, 32/20 = 8/5 → yes.

But 8/5 is acceptable. Alternatively, write as decimal? No, keep fraction.

Actually, 32/20 reduces to 8/5, yes.

✔ Final Answer: -3/10 + (8/5)i

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Problem 11: (6 - 4i) / (2 + 5i)

Multiply numerator and denominator by (2 - 5i)

Numerator: (6 - 4i)(2 - 5i) = 6*2 + 6*(-5i) + (-4i)*2 + (-4i)*(-5i)
= 12 -30i -8i +20i² = 12 -38i +20(-1) = 12 -38i -20 = -8 -38i

Denominator: (2 + 5i)(2 - 5i) = 4 - 25i² = 4 -25(-1) = 4 +25 = 29

So: (-8 -38i)/29 = -8/29 - (38/29)i

✔ Final Answer: -8/29 - (38/29)i

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Problem 12: (3 - 5i) / (5 + 2i)

Multiply numerator and denominator by (5 - 2i)

Numerator: (3 - 5i)(5 - 2i) = 3*5 + 3*(-2i) + (-5i)*5 + (-5i)*(-2i)
= 15 -6i -25i +10i² = 15 -31i +10(-1) = 15 -31i -10 = 5 -31i

Denominator: (5 + 2i)(5 - 2i) = 25 - 4i² = 25 -4(-1) = 25 +4 = 29

So: (5 -31i)/29 = 5/29 - (31/29)i

✔ Final Answer: 5/29 - (31/29)i

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Problem 13: (2 - 3i) / (3 - 3i) ← Wait, this is same as Problem 8!

In the original list, problem 8 was (2 - 3i)/(3 - 3i), and problem 13 is also (2 - 3i)/(3 - 3i)? That must be a typo in the worksheet? Or did I misread?

Looking back at user's image description:

Problem 8: (2 - 3i)/(3 - 3i)

Problem 13: (2 - 3i)/(3 - 3i) — same thing? Probably a copy-paste error in the worksheet.

But in my earlier solution for problem 8, I got 5/6 - (1/6)i

Let me recheck problem 8 quickly:

(2 - 3i)/(3 - 3i) * (3 + 3i)/(3 + 3i)

Numerator: (2-3i)(3+3i) = 6 +6i -9i -9i² = 6 -3i +9 = 15 -3i

Denominator: 9 - (3i)^2 = 9 -9i² = 9+9=18

15/18 - 3i/18 = 5/6 - 1/6 i → correct.

So problem 13 is identical? Maybe it’s a mistake. But since it’s listed again, perhaps it’s intentional? Or maybe I misread.

Wait — looking at original text:

"8. (2 - 3i) / (3 - 3i)"

"13. (2 - 3i) / (3 - 3i)" — yes, same.

Perhaps it’s a duplicate. I’ll assume it’s the same and give same answer.

But let’s check problem 14 too.

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Problem 14: (3 - 2i) / (-2i - 3i) — wait, that says “-2i - 3i”? That would be -5i.

Is that right? Let me read original:

"14. (3 - 2i) / (-2i - 3i)"

That seems odd. Denominator is -2i -3i = -5i.

So it’s (3 - 2i) / (-5i)

Then multiply numerator and denominator by i:

Numerator: (3 - 2i)i = 3i - 2i² = 3i +2

Denominator: -5i * i = -5i² = 5

So: (2 + 3i)/5 = 2/5 + (3/5)i

✔ Final Answer: 2/5 + (3/5)i

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Problem 15: (3 + 2i) / (2 - 4i)

Multiply numerator and denominator by (2 + 4i)

Numerator: (3 + 2i)(2 + 4i) = 3*2 + 3*4i + 2i*2 + 2i*4i = 6 + 12i + 4i + 8i²
= 6 + 16i + 8(-1) = 6 + 16i -8 = -2 + 16i

Denominator: (2 - 4i)(2 + 4i) = 4 - 16i² = 4 -16(-1) = 4 +16 = 20

So: (-2 + 16i)/20 = -2/20 + 16i/20 = -1/10 + (4/5)i

✔ Final Answer: -1/10 + (4/5)i

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Problem 16: (2 + 3i) / (2 - 4i)

Wait, this is similar to above but numerator different.

Multiply by (2 + 4i)

Numerator: (2 + 3i)(2 + 4i) = 2*2 + 2*4i + 3i*2 + 3i*4i = 4 + 8i + 6i + 12i²
= 4 + 14i + 12(-1) = 4 + 14i -12 = -8 + 14i

Denominator: same as before, 4 - 16i² = 4 +16 = 20

So: (-8 + 14i)/20 = -8/20 + 14i/20 = -2/5 + (7/10)i

✔ Final Answer: -2/5 + (7/10)i

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Problem 17: (3 - 2i) / (2 - 4i)

Multiply by (2 + 4i)

Numerator: (3 - 2i)(2 + 4i) = 3*2 + 3*4i -2i*2 -2i*4i = 6 + 12i -4i -8i²
= 6 + 8i -8(-1) = 6 + 8i +8 = 14 + 8i

Denominator: 20 (same as before)

So: (14 + 8i)/20 = 14/20 + 8i/20 = 7/10 + (2/5)i

✔ Final Answer: 7/10 + (2/5)i

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Problem 18: (2 - 3i) / (-2i - 3i) — again, denominator is -5i

Same as problem 14? Problem 14 was (3 - 2i)/(-5i), this is (2 - 3i)/(-5i)

So: (2 - 3i)/(-5i)

Multiply numerator and denominator by i:

Numerator: (2 - 3i)i = 2i - 3i² = 2i +3

Denominator: -5i * i = -5i² = 5

So: (3 + 2i)/5 = 3/5 + (2/5)i

✔ Final Answer: 3/5 + (2/5)i

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Problem 19: (3 - 2i) / (-2i - 3i) — again, denominator -5i

Same as above? Numerator is (3 - 2i), denominator -5i

So: (3 - 2i)/(-5i)

Multiply by i:

Numerator: (3 - 2i)i = 3i - 2i² = 3i +2

Denominator: -5i² = 5

So: (2 + 3i)/5 = 2/5 + (3/5)i

✔ Final Answer: 2/5 + (3/5)i

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Problem 20: (2 + 3i) / (2 - 4i) — this is same as problem 16!

Problem 16 was (2 + 3i)/(2 - 4i), and we got -2/5 + (7/10)i

So same answer.

✔ Final Answer: -2/5 + (7/10)i

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Now, compiling all answers in order:

1. -4i
2. 3i
3. -1/5 - (6/5)i
4. 42/53 + (12/53)i
5. 13/29 + (47/29)i
6. 4/3 + (5/3)i
7. 3 - i
8. 5/6 - (1/6)i
9. -2/29 + (24/29)i
10. -3/10 + (8/5)i
11. -8/29 - (38/29)i
12. 5/29 - (31/29)i
13. 5/6 - (1/6)i [duplicate of 8]
14. 2/5 + (3/5)i
15. -1/10 + (4/5)i
16. -2/5 + (7/10)i
17. 7/10 + (2/5)i
18. 3/5 + (2/5)i
19. 2/5 + (3/5)i [same as 14?]
20. -2/5 + (7/10)i [same as 16]

Note: Some problems are duplicates, likely due to worksheet error, but we solved them as given.

Final Answer:
1. -4i
2. 3i
3. -\frac{1}{5} - \frac{6}{5}i
4. \frac{42}{53} + \frac{12}{53}i
5. \frac{13}{29} + \frac{47}{29}i
6. \frac{4}{3} + \frac{5}{3}i
7. 3 - i
8. \frac{5}{6} - \frac{1}{6}i
9. -\frac{2}{29} + \frac{24}{29}i
10. -\frac{3}{10} + \frac{8}{5}i
11. -\frac{8}{29} - \frac{38}{29}i
12. \frac{5}{29} - \frac{31}{29}i
13. \frac{5}{6} - \frac{1}{6}i
14. \frac{2}{5} + \frac{3}{5}i
15. -\frac{1}{10} + \frac{4}{5}i
16. -\frac{2}{5} + \frac{7}{10}i
17. \frac{7}{10} + \frac{2}{5}i
18. \frac{3}{5} + \frac{2}{5}i
19. \frac{2}{5} + \frac{3}{5}i
20. -\frac{2}{5} + \frac{7}{10}i