Dividing Complex Numbers

To divide two complex numbers, $ a+bi $ and $ c+di $, we multiply both the numerator and denominator by the conjugate of the denominator, $ c-di $: $$ \frac{a+bi}{c+di} = \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} $$ Alternatively, we can use this formula. $$ \frac{a+bi}{c+di} = \frac{ac+bd}{c^2+d^2} + i \cdot \frac{bc-ad}{c^2+d^2} $$

In the trigonometric form, the division of two complex numbers, $ z_1 = m_1(\cos \gamma_1 + i \sin \gamma_1) $ and $ z_2 = m_2(\cos \gamma_2 + i \sin \gamma_2) $, follows this formula:

$$ \frac{z_1}{z_2} = \frac{m_1}{m_2} \cdot [ \cos (\gamma_1 - \gamma_2) + i \cdot \sin (\gamma_1 - \gamma_2) ] $$

Here, $ m $ represents the modulus, while $ \gamma $ is the argument (angle) of the complex number in the Gaussian plane.

For the exponential form, division is expressed as:

$$ \frac{z_1}{z_2} = \frac{m_1 \cdot e^{i\gamma_1 }}{m_2 \cdot e^{i\gamma_2}} = \left( \frac{m_1}{m_2} \right) \cdot e^{i(\gamma_1-\gamma_2)} $$

A Practical Example
Proof

A Practical Example

Let’s divide two complex numbers:

$$ z_1=3+2i $$ $$ z_2=4+7i $$

The division is:

$$ \frac{z_1}{z_2} = \frac{3+2i}{4+7i} $$

Method 1: Using the Conjugate

To simplify $ z_1 / z_2 $, we multiply both the numerator and denominator by the conjugate of $ z_2 $, which is $ z_2' = 4-7i $:

$$ \frac{z_1}{z_2} = \frac{3+2i}{4+7i} \cdot \frac{4-7i}{4-7i} $$

Expanding:

$$ = \frac{(3+2i) \cdot (4-7i)}{(4+7i) \cdot (4-7i)} $$

$$ = \frac{12 - 21i + 8i - 14i^2}{16 - 28i + 28i - 49i^2} $$

Since $ i^2 = -1 $, we simplify:

$$ = \frac{12 - 13i + 14}{16 + 49} $$

$$ = \frac{26 - 13i}{65} $$

$$ = \frac{26}{65} - \frac{13}{65}i $$

Method 2: Using the Division Formula

Alternatively, we can use the direct division formula for complex numbers:

$$ \frac{a+bi}{c+di} = \frac{ac+bd}{c^2+d^2} + i \cdot \frac{bc-ad}{c^2+d^2} $$

For $ z_1 = 3+2i $ and $ z_2 = 4+7i $, we apply the formula:

$$ \frac{(3 \cdot 4) + (2 \cdot 7)}{4^2 + 7^2} + i \cdot \frac{(4 \cdot 2) - (3 \cdot 7)}{4^2 + 7^2} $$

$$ = \frac{12+14}{16+49} + i \cdot \frac{8-21}{16+49} $$

$$ = \frac{26}{65} - i \cdot \frac{13}{65} $$

Both methods yield the same result.

Proof

To derive the division formula, we express division as multiplication.

$$ \frac{a+bi}{c+di} $$

Using the fundamental rule of division, we multiply both numerator and denominator by the conjugate of $ c+di $, which is $ c-di $:

$$ \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} $$

Note. The term $$ \frac{c-di}{(c+di) \cdot (c-di)} = \frac{c-di}{c^2+d^2} $$ is the reciprocal of $ c+di $, meaning: $$ \frac{c-di}{c^2+d^2} = \frac{1}{c+di} $$ This allows us to rewrite the expression as: $$ \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} = (a+bi) \cdot \frac{1}{c+di} $$ This confirms that dividing two complex numbers is equivalent to multiplying by the reciprocal of $ c+di $: $$ \frac{a+bi}{c+di} = (a+bi) \cdot \frac{1}{c+di} $$

Expanding:

$$ \frac{(a+bi) \cdot (c-di)}{(c+di) \cdot (c-di)} $$

$$ = \frac{ ac - adi + bci - bdi^2 }{ c^2 - d^2i^2 } $$

Since $ i^2 = -1 $, we simplify:

$$ = \frac{ ac - adi + bci + bd }{ c^2 + d^2 } $$

Factoring out $ i $:

$$ = \frac{ ac+ bd + (bc-ad)i }{ c^2 + d^2 } $$

And separating real and imaginary parts:

$$ = \frac{ ac+ bd }{ c^2 + d^2 } - \frac{ bc-ad }{ c^2 + d^2 } i $$

And that gives us the formula for dividing complex numbers.