Trigonometry Examples

${(2 - 2 i)}^{5}$

Step 1

Use the Binomial Theorem.

$2^{5} + 5 \cdot 2^{4} (- 2 i) + 10 \cdot 2^{3} {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2

Simplify terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

Raise $2$ to the power of $5$ .

$32 + 5 \cdot 2^{4} (- 2 i) + 10 \cdot 2^{3} {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.2

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

$32 + 5 \cdot 16 (- 2 i) + 10 \cdot 2^{3} {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.3

Multiply $5$ by $16$ .

$32 + 80 (- 2 i) + 10 \cdot 2^{3} {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.4

Multiply $- 2$ by $80$ .

$32 - 160 i + 10 \cdot 2^{3} {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.5

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

$32 - 160 i + 10 \cdot 8 {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.6

Multiply $10$ by $8$ .

$32 - 160 i + 80 {(- 2 i)}^{2} + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.7

Apply the product rule to $- 2 i$ .

$32 - 160 i + 80 ({(- 2)}^{2} i^{2}) + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.8

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

$32 - 160 i + 80 (4 i^{2}) + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.9

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

$32 - 160 i + 80 (4 \cdot - 1) + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.10

Multiply $80 (4 \cdot - 1)$ .

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Step 2.1.10.1

Multiply $4$ by $- 1$ .

$32 - 160 i + 80 \cdot - 4 + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.10.2

Multiply $80$ by $- 4$ .

$32 - 160 i - 320 + 10 \cdot 2^{2} {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.11

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

$32 - 160 i - 320 + 10 \cdot 4 {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.12

Multiply $10$ by $4$ .

$32 - 160 i - 320 + 40 {(- 2 i)}^{3} + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.13

Apply the product rule to $- 2 i$ .

$32 - 160 i - 320 + 40 ({(- 2)}^{3} i^{3}) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.14

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

$32 - 160 i - 320 + 40 (- 8 i^{3}) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.15

Factor out $i^{2}$ .

$32 - 160 i - 320 + 40 (- 8 (i^{2} \cdot i)) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.16

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

$32 - 160 i - 320 + 40 (- 8 (- 1 \cdot i)) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.17

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

$32 - 160 i - 320 + 40 (- 8 (- i)) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.18

Multiply $- 1$ by $- 8$ .

$32 - 160 i - 320 + 40 (8 i) + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.19

Multiply $8$ by $40$ .

$32 - 160 i - 320 + 320 i + 5 \cdot 2 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.20

Multiply $5$ by $2$ .

$32 - 160 i - 320 + 320 i + 10 {(- 2 i)}^{4} + {(- 2 i)}^{5}$

Step 2.1.21

Apply the product rule to $- 2 i$ .

$32 - 160 i - 320 + 320 i + 10 ({(- 2)}^{4} i^{4}) + {(- 2 i)}^{5}$

Step 2.1.22

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

$32 - 160 i - 320 + 320 i + 10 (16 i^{4}) + {(- 2 i)}^{5}$

Step 2.1.23

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

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Step 2.1.23.1

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

$32 - 160 i - 320 + 320 i + 10 (16 {(i^{2})}^{2}) + {(- 2 i)}^{5}$

Step 2.1.23.2

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

$32 - 160 i - 320 + 320 i + 10 (16 {(- 1)}^{2}) + {(- 2 i)}^{5}$

Step 2.1.23.3

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

$32 - 160 i - 320 + 320 i + 10 (16 \cdot 1) + {(- 2 i)}^{5}$

Step 2.1.24

Multiply $10 (16 \cdot 1)$ .

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Step 2.1.24.1

Multiply $16$ by $1$ .

$32 - 160 i - 320 + 320 i + 10 \cdot 16 + {(- 2 i)}^{5}$

Step 2.1.24.2

Multiply $10$ by $16$ .

$32 - 160 i - 320 + 320 i + 160 + {(- 2 i)}^{5}$

Step 2.1.25

Apply the product rule to $- 2 i$ .

$32 - 160 i - 320 + 320 i + 160 + {(- 2)}^{5} i^{5}$

Step 2.1.26

Raise $- 2$ to the power of $5$ .

$32 - 160 i - 320 + 320 i + 160 - 32 i^{5}$

Step 2.1.27

Factor out $i^{4}$ .

$32 - 160 i - 320 + 320 i + 160 - 32 (i^{4} i)$

Step 2.1.28

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

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Step 2.1.28.1

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

$32 - 160 i - 320 + 320 i + 160 - 32 ({(i^{2})}^{2} i)$

Step 2.1.28.2

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

$32 - 160 i - 320 + 320 i + 160 - 32 ({(- 1)}^{2} i)$

Step 2.1.28.3

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

$32 - 160 i - 320 + 320 i + 160 - 32 (1 i)$

Step 2.1.29

Multiply $i$ by $1$ .

$32 - 160 i - 320 + 320 i + 160 - 32 i$

Step 2.2

Simplify by adding terms.

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Step 2.2.1

Subtract $320$ from $32$ .

$- 288 - 160 i + 320 i + 160 - 32 i$

Step 2.2.2

Add $- 288$ and $160$ .

$- 128 - 160 i + 320 i - 32 i$

Step 2.2.3

Add $- 160 i$ and $320 i$ .

$- 128 + 160 i - 32 i$

Step 2.2.4

Subtract $32 i$ from $160 i$ .

$- 128 + 128 i$

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 $a = - 128$ and $b = 128$ .

$| z | = \sqrt{128^{2} + {(- 128)}^{2}}$

Step 6

Find $| z |$ .

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Step 6.1

Raise $128$ to the power of $2$ .

$| z | = \sqrt{16384 + {(- 128)}^{2}}$

Step 6.2

Raise $- 128$ to the power of $2$ .

$| z | = \sqrt{16384 + 16384}$

Step 6.3

Add $16384$ and $16384$ .

$| z | = \sqrt{32768}$

Step 6.4

Rewrite $32768$ as $128^{2} \cdot 2$ .

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Step 6.4.1

Factor $16384$ out of $32768$ .

$| z | = \sqrt{16384 (2)}$

Step 6.4.2

Rewrite $16384$ as $128^{2}$ .

$| z | = \sqrt{128^{2} \cdot 2}$

Step 6.5

Pull terms out from under the radical.

$| z | = 128 \sqrt{2}$

Step 7

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

$θ = \arctan (\frac{128}{- 128})$

Step 8

Since inverse tangent of $\frac{128}{- 128}$ produces an angle in the second quadrant, the value of the angle is $\frac{3 π}{4}$ .

$θ = \frac{3 π}{4}$

Step 9

Substitute the values of $θ = \frac{3 π}{4}$ and $| z | = 128 \sqrt{2}$ .

$128 \sqrt{2} (\cos (\frac{3 π}{4}) + i \sin (\frac{3 π}{4}))$