Rationalize Complex Number Worksheets (printable, online, answers ... - Free Printable

You're asked to simplify several complex number fractions. To do this, we use the technique of multiplying numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of a complex number $ a + bi $ is $ a - bi $. Multiplying a complex number by its conjugate gives a real number:
$ (a + bi)(a - bi) = a^2 + b^2 $

This eliminates the imaginary part from the denominator, allowing us to simplify the expression into standard form $ a + bi $.

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Let’s solve each one step-by-step.

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1. $ \frac{8 + 3i}{-3 + i} $

Step 1: Multiply numerator and denominator by the conjugate of the denominator: $ -3 - i $

$$
\frac{8 + 3i}{-3 + i} \cdot \frac{-3 - i}{-3 - i}
= \frac{(8 + 3i)(-3 - i)}{(-3 + i)(-3 - i)}
$$

Step 2: Compute denominator:

$$
(-3)^2 - (i)^2 = 9 - (-1) = 10
$$

Step 3: Compute numerator:

$$
(8)(-3) + (8)(-i) + (3i)(-3) + (3i)(-i)
= -24 -8i -9i -3i^2
= -24 -17i + 3 \quad (\text{since } i^2 = -1)
= -21 -17i
$$

Step 4: Put together:

$$
\frac{-21 -17i}{10} = -\frac{21}{10} - \frac{17}{10}i
$$

✔ Answer: $ -\frac{21}{10} - \frac{17}{10}i $

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2. $ \frac{9 + 7i}{1 - 3i} $

Multiply by conjugate $ 1 + 3i $:

$$
\frac{9 + 7i}{1 - 3i} \cdot \frac{1 + 3i}{1 + 3i}
= \frac{(9 + 7i)(1 + 3i)}{(1)^2 - (3i)^2}
= \frac{(9 + 7i)(1 + 3i)}{1 - (-9)} = \frac{(9 + 7i)(1 + 3i)}{10}
$$

Numerator:

$$
9(1) + 9(3i) + 7i(1) + 7i(3i)
= 9 + 27i + 7i + 21i^2
= 9 + 34i - 21
= -12 + 34i
$$

So,

$$
\frac{-12 + 34i}{10} = -\frac{6}{5} + \frac{17}{5}i
$$

✔ Answer: $ -\frac{6}{5} + \frac{17}{5}i $

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3. $ \frac{7 - 7i}{9 - i} $

Conjugate of denominator: $ 9 + i $

$$
\frac{7 - 7i}{9 - i} \cdot \frac{9 + i}{9 + i}
= \frac{(7 - 7i)(9 + i)}{81 - (-1)} = \frac{(7 - 7i)(9 + i)}{82}
$$

Numerator:

$$
7(9) + 7(i) -7i(9) -7i(i)
= 63 + 7i -63i -7i^2
= 63 -56i +7
= 70 -56i
$$

So,

$$
\frac{70 -56i}{82} = \frac{35 -28i}{41} \quad \text{(divide numerator and denominator by 2)}
$$

✔ Answer: $ \frac{35}{41} - \frac{28}{41}i $

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4. $ \frac{5 + i}{8 + i} $

Conjugate: $ 8 - i $

Denominator: $ 64 - (-1) = 65 $

Numerator:

$$
(5 + i)(8 - i) = 5(8) + 5(-i) + i(8) + i(-i)
= 40 -5i +8i -i^2
= 40 +3i +1
= 41 +3i
$$

So,

$$
\frac{41 +3i}{65} = \frac{41}{65} + \frac{3}{65}i
$$

✔ Answer: $ \frac{41}{65} + \frac{3}{65}i $

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5. $ \frac{-10 + i}{-1 + 3i} $

Conjugate: $ -1 - 3i $

Denominator: $ (-1)^2 - (3i)^2 = 1 - (-9) = 10 $

Numerator:

$$
(-10 + i)(-1 - 3i)
= (-10)(-1) + (-10)(-3i) + i(-1) + i(-3i)
= 10 + 30i -i -3i^2
= 10 + 29i +3
= 13 + 29i
$$

So,

$$
\frac{13 + 29i}{10} = \frac{13}{10} + \frac{29}{10}i
$$

✔ Answer: $ \frac{13}{10} + \frac{29}{10}i $

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6. $ \frac{-3 -5i}{-8 + 3i} $

Conjugate: $ -8 - 3i $

Denominator: $ (-8)^2 - (3i)^2 = 64 - (-9) = 73 $

Numerator:

$$
(-3 -5i)(-8 -3i)
= (-3)(-8) + (-3)(-3i) + (-5i)(-8) + (-5i)(-3i)
= 24 + 9i + 40i + 15i^2
= 24 + 49i -15
= 9 + 49i
$$

So,

$$
\frac{9 + 49i}{73} = \frac{9}{73} + \frac{49}{73}i
$$

✔ Answer: $ \frac{9}{73} + \frac{49}{73}i $

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7. $ \frac{1 + 2i}{5 + i} $

Conjugate: $ 5 - i $

Denominator: $ 25 - (-1) = 26 $

Numerator:

$$
(1 + 2i)(5 - i) = 1(5) + 1(-i) + 2i(5) + 2i(-i)
= 5 -i +10i -2i^2
= 5 +9i +2
= 7 +9i
$$

So,

$$
\frac{7 +9i}{26} = \frac{7}{26} + \frac{9}{26}i
$$

✔ Answer: $ \frac{7}{26} + \frac{9}{26}i $

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8. $ \frac{3 + 3i}{9 - i} $

Conjugate: $ 9 + i $

Denominator: $ 81 - (-1) = 82 $

Numerator:

$$
(3 + 3i)(9 + i) = 3(9) + 3(i) + 3i(9) + 3i(i)
= 27 + 3i + 27i + 3i^2
= 27 + 30i -3
= 24 + 30i
$$

So,

$$
\frac{24 + 30i}{82} = \frac{12 + 15i}{41} \quad \text{(divide by 2)}
$$

✔ Answer: $ \frac{12}{41} + \frac{15}{41}i $

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## ✔ Final Answers:

1. $ \boxed{-\frac{21}{10} - \frac{17}{10}i} $
2. $ \boxed{-\frac{6}{5} + \frac{17}{5}i} $
3. $ \boxed{\frac{35}{41} - \frac{28}{41}i} $
4. $ \boxed{\frac{41}{65} + \frac{3}{65}i} $
5. $ \boxed{\frac{13}{10} + \frac{29}{10}i} $
6. $ \boxed{\frac{9}{73} + \frac{49}{73}i} $
7. $ \boxed{\frac{7}{26} + \frac{9}{26}i} $
8. $ \boxed{\frac{12}{41} + \frac{15}{41}i} $

Let me know if you’d like to see any step expanded further!