Linear Algebra Examples

3 i - 2

$3 i - 2$

Step 1

Reorder

3 i

$3 i$ and

- 2

$- 2$ .

- 2 + 3 i

$- 2 + 3 i$

Step 2

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 3

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 4

Substitute the actual values of

a = - 2

$a = - 2$ and

b = 3

$b = 3$ .

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

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

Step 5

Find

| z |

$| z |$ .

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Step 5.1

Raise

3

$3$ to the power of

2

$2$ .

| z | = \sqrt{9 + {(- 2)}^{2}}

$| z | = \sqrt{9 + {(- 2)}^{2}}$

Step 5.2

Raise

- 2

$- 2$ to the power of

2

$2$ .

| z | = \sqrt{9 + 4}

$| z | = \sqrt{9 + 4}$

Step 5.3

Add

9

$9$ and

4

$4$ .

| z | = \sqrt{13}

$| z | = \sqrt{13}$

| z | = \sqrt{13}

$| z | = \sqrt{13}$

Step 6

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

θ = arctan (\frac{3}{- 2})

$θ = \arctan (\frac{3}{- 2})$

Step 7

Since inverse tangent of

\frac{3}{- 2}

$\frac{3}{- 2}$ produces an angle in the second quadrant, the value of the angle is

2.15879893

$2.15879893$ .

θ = 2.15879893

$θ = 2.15879893$

Step 8

Substitute the values of

θ = 2.15879893

$θ = 2.15879893$ and

| z | = \sqrt{13}

$| z | = \sqrt{13}$ .

\sqrt{13} (cos (2.15879893) + i sin (2.15879893))

$\sqrt{13} (\cos (2.15879893) + i \sin (2.15879893))$

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