Linear Algebra Examples

$(- 3 - 5 i) - (4 + 2 i)$

Step 1

Simplify each term.

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Step 1.1

Apply the distributive property.

$- 3 - 5 i - 1 \cdot 4 - (2 i)$

Step 1.2

Multiply $- 1$ by $4$ .

$- 3 - 5 i - 4 - (2 i)$

Step 1.3

Multiply $2$ by $- 1$ .

$- 3 - 5 i - 4 - 2 i$

Step 2

Simplify by adding terms.

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Step 2.1

Subtract $4$ from $- 3$ .

$- 7 - 5 i - 2 i$

Step 2.2

Subtract $2 i$ from $- 5 i$ .

$- 7 - 7 i$

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 $a = - 7$ and $b = - 7$ .

$| z | = \sqrt{{(- 7)}^{2} + {(- 7)}^{2}}$

Step 6

Find $| z |$ .

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Step 6.1

Raise $- 7$ to the power of $2$ .

$| z | = \sqrt{49 + {(- 7)}^{2}}$

Step 6.2

Raise $- 7$ to the power of $2$ .

$| z | = \sqrt{49 + 49}$

Step 6.3

Add $49$ and $49$ .

$| z | = \sqrt{98}$

Step 6.4

Rewrite $98$ as $7^{2} \cdot 2$ .

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Step 6.4.1

Factor $49$ out of $98$ .

$| z | = \sqrt{49 (2)}$

Step 6.4.2

Rewrite $49$ as $7^{2}$ .

$| z | = \sqrt{7^{2} \cdot 2}$

Step 6.5

Pull terms out from under the radical.

$| z | = 7 \sqrt{2}$

Step 7

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

$θ = \arctan (\frac{- 7}{- 7})$

Step 8

Since inverse tangent of $\frac{- 7}{- 7}$ produces an angle in the third quadrant, the value of the angle is $\frac{5 π}{4}$ .

$θ = \frac{5 π}{4}$

Step 9

Substitute the values of $θ = \frac{5 π}{4}$ and $| z | = 7 \sqrt{2}$ .

$7 \sqrt{2} (\cos (\frac{5 π}{4}) + i \sin (\frac{5 π}{4}))$