Precalculus Examples

$- 4 \sqrt{2} (1 - i)$

Step 1

Apply the distributive property.

$- 4 \sqrt{2} \cdot 1 - 4 \sqrt{2} (- i)$

Step 2

Multiply.

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Step 2.1

Multiply $- 4$ by $1$ .

$- 4 \sqrt{2} - 4 \sqrt{2} (- i)$

Step 2.2

Multiply $- 1$ by $- 4$ .

$- 4 \sqrt{2} + 4 \sqrt{2} i$

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 $a = - 4 \sqrt{2}$ and $b = 4 \sqrt{2}$ .

$| z | = \sqrt{{(4 \sqrt{2})}^{2} + {(- 4 \sqrt{2})}^{2}}$

Step 6

Find $| z |$ .

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Step 6.1

Simplify the expression.

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Step 6.1.1

Apply the product rule to $4 \sqrt{2}$ .

$| z | = \sqrt{4^{2} {\sqrt{2}}^{2} + {(- 4 \sqrt{2})}^{2}}$

Step 6.1.2

Raise $4$ to the power of $2$ .

$| z | = \sqrt{16 {\sqrt{2}}^{2} + {(- 4 \sqrt{2})}^{2}}$

Step 6.2

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 6.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$| z | = \sqrt{16 {(2^{\frac{1}{2}})}^{2} + {(- 4 \sqrt{2})}^{2}}$

Step 6.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| z | = \sqrt{16 \cdot 2^{\frac{1}{2} \cdot 2} + {(- 4 \sqrt{2})}^{2}}$

Step 6.2.3

Combine $\frac{1}{2}$ and $2$ .

$| z | = \sqrt{16 \cdot 2^{\frac{2}{2}} + {(- 4 \sqrt{2})}^{2}}$

Step 6.2.4

Cancel the common factor of $2$ .

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Step 6.2.4.1

Cancel the common factor.

$| z | = \sqrt{16 \cdot 2^{\frac{2}{2}} + {(- 4 \sqrt{2})}^{2}}$

Step 6.2.4.2

Rewrite the expression.

$| z | = \sqrt{16 \cdot 2 + {(- 4 \sqrt{2})}^{2}}$

Step 6.2.5

Evaluate the exponent.

$| z | = \sqrt{16 \cdot 2 + {(- 4 \sqrt{2})}^{2}}$

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $16$ by $2$ .

$| z | = \sqrt{32 + {(- 4 \sqrt{2})}^{2}}$

Step 6.3.2

Apply the product rule to $- 4 \sqrt{2}$ .

$| z | = \sqrt{32 + {(- 4)}^{2} {\sqrt{2}}^{2}}$

Step 6.3.3

Raise $- 4$ to the power of $2$ .

$| z | = \sqrt{32 + 16 {\sqrt{2}}^{2}}$

Step 6.4

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 6.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$| z | = \sqrt{32 + 16 {(2^{\frac{1}{2}})}^{2}}$

Step 6.4.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| z | = \sqrt{32 + 16 \cdot 2^{\frac{1}{2} \cdot 2}}$

Step 6.4.3

Combine $\frac{1}{2}$ and $2$ .

$| z | = \sqrt{32 + 16 \cdot 2^{\frac{2}{2}}}$

Step 6.4.4

Cancel the common factor of $2$ .

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Step 6.4.4.1

Cancel the common factor.

$| z | = \sqrt{32 + 16 \cdot 2^{\frac{2}{2}}}$

Step 6.4.4.2

Rewrite the expression.

$| z | = \sqrt{32 + 16 \cdot 2}$

Step 6.4.5

Evaluate the exponent.

$| z | = \sqrt{32 + 16 \cdot 2}$

Step 6.5

Simplify the expression.

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Step 6.5.1

Multiply $16$ by $2$ .

$| z | = \sqrt{32 + 32}$

Step 6.5.2

Add $32$ and $32$ .

$| z | = \sqrt{64}$

Step 6.5.3

Rewrite $64$ as $8^{2}$ .

$| z | = \sqrt{8^{2}}$

Step 6.5.4

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

$| z | = 8$

Step 7

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

$θ = \arctan (\frac{4 \sqrt{2}}{- 4 \sqrt{2}})$

Step 8

Since inverse tangent of $\frac{4 \sqrt{2}}{- 4 \sqrt{2}}$ produces an angle in the second quadrant, the value of the angle is $\frac{3 π}{4}$ .

$θ = \frac{3 π}{4}$

Step 9

Substitute the values of $θ = \frac{3 π}{4}$ and $| z | = 8$ .

$8 (\cos (\frac{3 π}{4}) + i \sin (\frac{3 π}{4}))$