Trigonometry Examples

$1 - \sqrt{3} i$

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 $a = 1$ and $b = - 1 \sqrt{3}$ .

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

Step 4

Find $| z |$ .

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Step 4.1

Simplify the expression.

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Step 4.1.1

Rewrite $- 1 \sqrt{3}$ as $- \sqrt{3}$ .

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

Step 4.1.2

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

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

Step 4.1.3

Raise $- 1$ to the power of $2$ .

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

Step 4.1.4

Multiply ${\sqrt{3}}^{2}$ by $1$ .

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

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 $3^{\frac{1}{2}}$ .

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

Step 4.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Cancel the common factor of $2$ .

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Step 4.2.4.1

Cancel the common factor.

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

Step 4.2.4.2

Rewrite the expression.

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

Step 4.2.5

Evaluate the exponent.

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

Step 4.3

Simplify the expression.

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Step 4.3.1

One to any power is one.

$| z | = \sqrt{3 + 1}$

Step 4.3.2

Add $3$ and $1$ .

$| z | = \sqrt{4}$

Step 4.3.3

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

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

Step 4.4

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

$| z | = 2$

Step 5

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

$θ = \arctan (\frac{- 1 \sqrt{3}}{1})$

Step 6

Since inverse tangent of $\frac{- 1 \sqrt{3}}{1}$ produces an angle in the fourth quadrant, the value of the angle is $- \frac{π}{3}$ .

$θ = - \frac{π}{3}$

Step 7

Substitute the values of $θ = - \frac{π}{3}$ and $| z | = 2$ .

$2 (\cos (- \frac{π}{3}) + i \sin (- \frac{π}{3}))$