Trigonometry Examples

${(1 + i)}^{4}$

Step 1

Use the Binomial Theorem.

$1^{4} + 4 \cdot 1^{3} i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2

Simplify terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

One to any power is one.

$1 + 4 \cdot 1^{3} i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.2

One to any power is one.

$1 + 4 \cdot 1 i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.3

Multiply $4$ by $1$ .

$1 + 4 i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.4

One to any power is one.

$1 + 4 i + 6 \cdot 1 i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.5

Multiply $6$ by $1$ .

$1 + 4 i + 6 i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.6

Rewrite $i^{2}$ as $- 1$ .

$1 + 4 i + 6 \cdot - 1 + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.7

Multiply $6$ by $- 1$ .

$1 + 4 i - 6 + 4 \cdot 1 i^{3} + i^{4}$

Step 2.1.8

Multiply $4$ by $1$ .

$1 + 4 i - 6 + 4 i^{3} + i^{4}$

Step 2.1.9

Factor out $i^{2}$ .

$1 + 4 i - 6 + 4 (i^{2} \cdot i) + i^{4}$

Step 2.1.10

Rewrite $i^{2}$ as $- 1$ .

$1 + 4 i - 6 + 4 (- 1 \cdot i) + i^{4}$

Step 2.1.11

Rewrite $- 1 i$ as $- i$ .

$1 + 4 i - 6 + 4 (- i) + i^{4}$

Step 2.1.12

Multiply $- 1$ by $4$ .

$1 + 4 i - 6 - 4 i + i^{4}$

Step 2.1.13

Rewrite $i^{4}$ as $1$ .

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Step 2.1.13.1

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$1 + 4 i - 6 - 4 i + {(i^{2})}^{2}$

Step 2.1.13.2

Rewrite $i^{2}$ as $- 1$ .

$1 + 4 i - 6 - 4 i + {(- 1)}^{2}$

Step 2.1.13.3

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

$1 + 4 i - 6 - 4 i + 1$

Step 2.2

Simplify by adding terms.

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Step 2.2.1

Subtract $6$ from $1$ .

$- 5 + 4 i - 4 i + 1$

Step 2.2.2

Add $- 5$ and $1$ .

$- 4 + 4 i - 4 i$

Step 2.2.3

Subtract $4 i$ from $4 i$ .

$- 4 + 0$

Step 2.2.4

Add $- 4$ and $0$ .

$- 4$

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 $a = - 4$ and $b = 0$ .

$| z | = \sqrt{0^{2} + {(- 4)}^{2}}$

Step 6

Find $| z |$ .

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Step 6.1

Raising $0$ to any positive power yields $0$ .

$| z | = \sqrt{0 + {(- 4)}^{2}}$

Step 6.2

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

$| z | = \sqrt{0 + 16}$

Step 6.3

Add $0$ and $16$ .

$| z | = \sqrt{16}$

Step 6.4

Rewrite $16$ as $4^{2}$ .

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

Step 6.5

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

$| z | = 4$

Step 7

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

$θ = \arctan (\frac{0}{- 4})$

Step 8

Since inverse tangent of $\frac{0}{- 4}$ produces an angle in the second quadrant, the value of the angle is $π$ .

$θ = π$

Step 9

Substitute the values of $θ = π$ and $| z | = 4$ .

$4 (\cos (π) + i \sin (π))$