Linear Algebra Examples

$(12 - 8 i) - (6 - 6 i)$

Step 1

Simplify each term.

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Step 1.1

Apply the distributive property.

$12 - 8 i - 1 \cdot 6 - (- 6 i)$

Step 1.2

Multiply $- 1$ by $6$ .

$12 - 8 i - 6 - (- 6 i)$

Step 1.3

Multiply $- 6$ by $- 1$ .

$12 - 8 i - 6 + 6 i$

Step 2

Simplify by adding terms.

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Step 2.1

Subtract $6$ from $12$ .

$6 - 8 i + 6 i$

Step 2.2

Add $- 8 i$ and $6 i$ .

$6 - 2 i$

Step 3

This is the trigonometric form of a complex number where $| z |$ is the modulus and $θ$ is the angle created on the complex plane.

$z = a + b i = | z | (\cos (θ) + i \sin (θ))$

Step 4

The modulus of a complex number is the distance from the origin on the complex plane.

$| z | = \sqrt{a^{2} + b^{2}}$ where $z = a + b i$

Step 5

Substitute the actual values of $a = 6$ and $b = - 2$ .

$| z | = \sqrt{{(- 2)}^{2} + 6^{2}}$

Step 6

Find $| z |$ .

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Step 6.1

Raise $- 2$ to the power of $2$ .

$| z | = \sqrt{4 + 6^{2}}$

Step 6.2

Raise $6$ to the power of $2$ .

$| z | = \sqrt{4 + 36}$

Step 6.3

Add $4$ and $36$ .

$| z | = \sqrt{40}$

Step 6.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 6.4.1

Factor $4$ out of $40$ .

$| z | = \sqrt{4 (10)}$

Step 6.4.2

Rewrite $4$ as $2^{2}$ .

$| z | = \sqrt{2^{2} \cdot 10}$

Step 6.5

Pull terms out from under the radical.

$| z | = 2 \sqrt{10}$

Step 7

The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion.

$θ = \arctan (\frac{- 2}{6})$

Step 8

Since inverse tangent of $\frac{- 2}{6}$ produces an angle in the fourth quadrant, the value of the angle is $- 0.32175055$ .

$θ = - 0.32175055$

Step 9

Substitute the values of $θ = - 0.32175055$ and $| z | = 2 \sqrt{10}$ .

$2 \sqrt{10} (\cos (- 0.32175055) + i \sin (- 0.32175055))$