Linear Algebra Examples

$- 5 i {(4 - 3 i)}^{2}$

Step 1

Rewrite ${(4 - 3 i)}^{2}$ as $(4 - 3 i) (4 - 3 i)$ .

$- 5 i ((4 - 3 i) (4 - 3 i))$

Step 2

Expand $(4 - 3 i) (4 - 3 i)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

$- 5 i (4 (4 - 3 i) - 3 i (4 - 3 i))$

Step 2.2

Apply the distributive property.

$- 5 i (4 \cdot 4 + 4 (- 3 i) - 3 i (4 - 3 i))$

Step 2.3

Apply the distributive property.

$- 5 i (4 \cdot 4 + 4 (- 3 i) - 3 i \cdot 4 - 3 i (- 3 i))$

Step 3

Simplify and combine like terms.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $4$ by $4$ .

$- 5 i (16 + 4 (- 3 i) - 3 i \cdot 4 - 3 i (- 3 i))$

Step 3.1.2

Multiply $- 3$ by $4$ .

$- 5 i (16 - 12 i - 3 i \cdot 4 - 3 i (- 3 i))$

Step 3.1.3

Multiply $4$ by $- 3$ .

$- 5 i (16 - 12 i - 12 i - 3 i (- 3 i))$

Step 3.1.4

Multiply $- 3 i (- 3 i)$ .

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Step 3.1.4.1

Multiply $- 3$ by $- 3$ .

$- 5 i (16 - 12 i - 12 i + 9 i i)$

Step 3.1.4.2

Raise $i$ to the power of $1$ .

$- 5 i (16 - 12 i - 12 i + 9 (i^{1} i))$

Step 3.1.4.3

Raise $i$ to the power of $1$ .

$- 5 i (16 - 12 i - 12 i + 9 (i^{1} i^{1}))$

Step 3.1.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 5 i (16 - 12 i - 12 i + 9 i^{1 + 1})$

Step 3.1.4.5

Add $1$ and $1$ .

$- 5 i (16 - 12 i - 12 i + 9 i^{2})$

Step 3.1.5

Rewrite $i^{2}$ as $- 1$ .

$- 5 i (16 - 12 i - 12 i + 9 \cdot - 1)$

Step 3.1.6

Multiply $9$ by $- 1$ .

$- 5 i (16 - 12 i - 12 i - 9)$

Step 3.2

Subtract $9$ from $16$ .

$- 5 i (7 - 12 i - 12 i)$

Step 3.3

Subtract $12 i$ from $- 12 i$ .

$- 5 i (7 - 24 i)$

Step 4

Apply the distributive property.

$- 5 i \cdot 7 - 5 i (- 24 i)$

Step 5

Multiply $7$ by $- 5$ .

$- 35 i - 5 i (- 24 i)$

Step 6

Multiply $- 5 i (- 24 i)$ .

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Step 6.1

Multiply $- 24$ by $- 5$ .

$- 35 i + 120 i i$

Step 6.2

Raise $i$ to the power of $1$ .

$- 35 i + 120 (i^{1} i)$

Step 6.3

Raise $i$ to the power of $1$ .

$- 35 i + 120 (i^{1} i^{1})$

Step 6.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 35 i + 120 i^{1 + 1}$

Step 6.5

Add $1$ and $1$ .

$- 35 i + 120 i^{2}$

Step 7

Simplify each term.

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Step 7.1

Rewrite $i^{2}$ as $- 1$ .

$- 35 i + 120 \cdot - 1$

Step 7.2

Multiply $120$ by $- 1$ .

$- 35 i - 120$

Step 8

Reorder $- 35 i$ and $- 120$ .

$- 120 - 35 i$

Step 9

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 10

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 11

Substitute the actual values of $a = - 120$ and $b = - 35$ .

$| z | = \sqrt{{(- 35)}^{2} + {(- 120)}^{2}}$

Step 12

Find $| z |$ .

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Step 12.1

Raise $- 35$ to the power of $2$ .

$| z | = \sqrt{1225 + {(- 120)}^{2}}$

Step 12.2

Raise $- 120$ to the power of $2$ .

$| z | = \sqrt{1225 + 14400}$

Step 12.3

Add $1225$ and $14400$ .

$| z | = \sqrt{15625}$

Step 12.4

Rewrite $15625$ as $125^{2}$ .

$| z | = \sqrt{125^{2}}$

Step 12.5

Pull terms out from under the radical, assuming positive real numbers.

$| z | = 125$

Step 13

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

$θ = \arctan (\frac{- 35}{- 120})$

Step 14

Since inverse tangent of $\frac{- 35}{- 120}$ produces an angle in the third quadrant, the value of the angle is $3.42538676$ .

$θ = 3.42538676$

Step 15

Substitute the values of $θ = 3.42538676$ and $| z | = 125$ .

$125 (\cos (3.42538676) + i \sin (3.42538676))$