Trigonometry Examples

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

Step 1

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 2

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 3

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

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

Step 4

Find $| z |$ .

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Step 4.1

Simplify the expression.

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Step 4.1.1

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

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

Step 4.1.2

Raise $- 3$ to the power of $2$ .

$| z | = \sqrt{9 {\sqrt{3}}^{2} + {(- 3)}^{2}}$

Step 4.2

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

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Step 4.2.1

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

$| z | = \sqrt{9 {(3^{\frac{1}{2}})}^{2} + {(- 3)}^{2}}$

Step 4.2.2

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

$| z | = \sqrt{9 \cdot 3^{\frac{1}{2} \cdot 2} + {(- 3)}^{2}}$

Step 4.2.3

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

$| z | = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(- 3)}^{2}}$

Step 4.2.4

Cancel the common factor of $2$ .

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Step 4.2.4.1

Cancel the common factor.

$| z | = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(- 3)}^{2}}$

Step 4.2.4.2

Rewrite the expression.

$| z | = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

Step 4.2.5

Evaluate the exponent.

$| z | = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $9$ by $3$ .

$| z | = \sqrt{27 + {(- 3)}^{2}}$

Step 4.3.2

Raise $- 3$ to the power of $2$ .

$| z | = \sqrt{27 + 9}$

Step 4.3.3

Add $27$ and $9$ .

$| z | = \sqrt{36}$

Step 4.3.4

Rewrite $36$ as $6^{2}$ .

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

Step 4.3.5

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

$| z | = 6$

Step 5

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

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

Step 6

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

$θ = \frac{4 π}{3}$

Step 7

Substitute the values of $θ = \frac{4 π}{3}$ and $| z | = 6$ .

$6 (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$