Precalculus Examples

$- 5 + 2 i$

Step 1

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 2

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 3

Substitute the actual values of $a = - 5$ and $b = 2$ .

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

Step 4

Find $| z |$ .

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Step 4.1

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

$| z | = \sqrt{4 + {(- 5)}^{2}}$

Step 4.2

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

$| z | = \sqrt{4 + 25}$

Step 4.3

Add $4$ and $25$ .

$| z | = \sqrt{29}$

Step 5

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

$θ = \arctan (\frac{2}{- 5})$

Step 6

Since inverse tangent of $\frac{2}{- 5}$ produces an angle in the second quadrant, the value of the angle is $2.76108627$ .

$θ = 2.76108627$

Step 7

Substitute the values of $θ = 2.76108627$ and $| z | = \sqrt{29}$ .

$\sqrt{29} (\cos (2.76108627) + i \sin (2.76108627))$