Linear Algebra Examples

3 - 5 i

$3 - 5 i$

Step 1

This is the trigonometric form of a complex number where

| z |

$| z |$ is the modulus and

θ

$θ$ is the angle created on the complex plane.

z = a + b i = | z | (\cos (θ) + i \sin (θ))

$z = a + b i = | z | (\cos (θ) + i \sin (θ))$

Step 2

The modulus of a complex number is the distance from the origin on the complex plane.

| z | = \sqrt{a^{2} + b^{2}}

$| z | = \sqrt{a^{2} + b^{2}}$ where

z = a + b i

$z = a + b i$

Step 3

Substitute the actual values of

a = 3

$a = 3$ and

b = - 5

$b = - 5$ .

| z | = \sqrt{{(- 5)}^{2} + 3^{2}}

$| z | = \sqrt{{(- 5)}^{2} + 3^{2}}$

Step 4

Find

| z |

$| z |$ .

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Step 4.1

Raise

- 5

$- 5$ to the power of

2

$2$ .

| z | = \sqrt{25 + 3^{2}}

$| z | = \sqrt{25 + 3^{2}}$

Step 4.2

Raise

3

$3$ to the power of

2

$2$ .

| z | = \sqrt{25 + 9}

$| z | = \sqrt{25 + 9}$

Step 4.3

Add

25

$25$ and

9

$9$ .

| z | = \sqrt{34}

$| z | = \sqrt{34}$

| z | = \sqrt{34}

$| z | = \sqrt{34}$

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{- 5}{3})

$θ = \arctan (\frac{- 5}{3})$

Step 6

Since inverse tangent of

\frac{- 5}{3}

$\frac{- 5}{3}$ produces an angle in the fourth quadrant, the value of the angle is

- 1.03037682

$- 1.03037682$ .

θ = - 1.03037682

$θ = - 1.03037682$

Step 7

Substitute the values of

θ = - 1.03037682

$θ = - 1.03037682$ and

| z | = \sqrt{34}

$| z | = \sqrt{34}$ .

\sqrt{34} (\cos (- 1.03037682) + i \sin (- 1.03037682))

$\sqrt{34} (\cos (- 1.03037682) + i \sin (- 1.03037682))$