Precalculus Examples

$4 + 4 i \sqrt{3}$

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 $a = 4$ and $b = 4 \sqrt{3}$ .

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

Step 4

Find $| z |$ .

Tap for more steps...

Step 4.1

Simplify the expression.

Tap for more steps...

Step 4.1.1

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

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

Step 4.1.2

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

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

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

$| z | = \sqrt{16 {(3^{\frac{1}{2}})}^{2} + 4^{2}}$

Step 4.2.2

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

$| z | = \sqrt{16 \cdot 3^{\frac{1}{2} \cdot 2} + 4^{2}}$

Step 4.2.3

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

$| z | = \sqrt{16 \cdot 3^{\frac{2}{2}} + 4^{2}}$

Step 4.2.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.2.4.1

Cancel the common factor.

$| z | = \sqrt{16 \cdot 3^{\frac{2}{2}} + 4^{2}}$

Step 4.2.4.2

Rewrite the expression.

$| z | = \sqrt{16 \cdot 3 + 4^{2}}$

Step 4.2.5

Evaluate the exponent.

$| z | = \sqrt{16 \cdot 3 + 4^{2}}$

Step 4.3

Simplify the expression.

Tap for more steps...

Step 4.3.1

Multiply $16$ by $3$ .

$| z | = \sqrt{48 + 4^{2}}$

Step 4.3.2

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

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

Step 4.3.3

Add $48$ and $16$ .

$| z | = \sqrt{64}$

Step 4.3.4

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

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

Step 4.3.5

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

$| z | = 8$

Step 5

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

$θ = \arctan (\frac{4 \sqrt{3}}{4})$

Step 6

Since inverse tangent of $\frac{4 \sqrt{3}}{4}$ produces an angle in the first 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 $| z | = 8$ .

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