Trigonometry Examples

$- 4 - 4 i \sqrt{3} + 4 i - 4 \sqrt{3}$

Step 1

Reorder $- 4 - 4 i \sqrt{3} + 4 i$ and $- 4 \sqrt{3}$ .

$- 4 \sqrt{3} - 4 - 4 i \sqrt{3} + 4 i$

Step 2

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 3

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 4

Substitute the actual values of $a = - 4 \sqrt{3}$ and $b = - 4$ .

$| z | = \sqrt{{(- 4)}^{2} + {(- 4 \sqrt{3})}^{2}}$

Step 5

Find $| z |$ .

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Step 5.1

Simplify the expression.

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Step 5.1.1

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

$| z | = \sqrt{16 + {(- 4 \sqrt{3})}^{2}}$

Step 5.1.2

Apply the product rule to $- 4 \sqrt{3}$ .

$| z | = \sqrt{16 + {(- 4)}^{2} {\sqrt{3}}^{2}}$

Step 5.1.3

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

$| z | = \sqrt{16 + 16 {\sqrt{3}}^{2}}$

Step 5.2

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 5.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$| z | = \sqrt{16 + 16 {(3^{\frac{1}{2}})}^{2}}$

Step 5.2.2

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

$| z | = \sqrt{16 + 16 \cdot 3^{\frac{1}{2} \cdot 2}}$

Step 5.2.3

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

$| z | = \sqrt{16 + 16 \cdot 3^{\frac{2}{2}}}$

Step 5.2.4

Cancel the common factor of $2$ .

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Step 5.2.4.1

Cancel the common factor.

$| z | = \sqrt{16 + 16 \cdot 3^{\frac{2}{2}}}$

Step 5.2.4.2

Rewrite the expression.

$| z | = \sqrt{16 + 16 \cdot 3}$

Step 5.2.5

Evaluate the exponent.

$| z | = \sqrt{16 + 16 \cdot 3}$

Step 5.3

Simplify the expression.

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Step 5.3.1

Multiply $16$ by $3$ .

$| z | = \sqrt{16 + 48}$

Step 5.3.2

Add $16$ and $48$ .

$| z | = \sqrt{64}$

Step 5.3.3

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

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

Step 5.3.4

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

$| z | = 8$

Step 6

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

$θ = \arctan (\frac{- 4}{- 4 \sqrt{3}})$

Step 7

Since inverse tangent of $\frac{- 4}{- 4 \sqrt{3}}$ produces an angle in the third quadrant, the value of the angle is $\frac{7 π}{6}$ .

$θ = \frac{7 π}{6}$

Step 8

Substitute the values of $θ = \frac{7 π}{6}$ and $| z | = 8$ .

$8 (\cos (\frac{7 π}{6}) + i \sin (\frac{7 π}{6}))$