Convert to Trigonometric Form 3(cos(pi)+isin(pi))

$3 (\cos (π) + i \sin (π))$

Simplify each term.

Tap for more steps...

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$3 (- \cos (0) + i \sin (π))$

The exact value of $\cos (0)$ is $1$ .

$3 (- 1 \cdot 1 + i \sin (π))$

Multiply $- 1$ by $1$ .

$3 (- 1 + i \sin (π))$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$3 (- 1 + i \sin (0))$

The exact value of $\sin (0)$ is $0$ .

$3 (- 1 + i \cdot 0)$

Multiply $i$ by $0$ .

$3 (- 1 + 0)$

Simplify the expression.

Tap for more steps...

Add $- 1$ and $0$ .

$3 \cdot - 1$

Multiply $3$ by $- 1$ .

$- 3$

This is the trigonometric form of a complex number where $| z |$ is the modulus and $θ$ is the angle created on the complex plane.

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

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

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

Substitute the actual values of $a = - 3$ and $b = 0$ .

$| z | = \sqrt{0^{2} + {(- 3)}^{2}}$

Find $| z |$ .

Tap for more steps...

Raising $0$ to any positive power yields $0$ .

$| z | = \sqrt{0 + {(- 3)}^{2}}$

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

$| z | = \sqrt{0 + 9}$

Add $0$ and $9$ .

$| z | = \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

$| z | = 3$

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

$θ = \arctan (\frac{0}{- 3})$

Since inverse tangent of $\frac{0}{- 3}$ produces an angle in the second quadrant, the value of the angle is $π$ .

$θ = π$

Substitute the values of $θ = π$ and $| z | = 3$ .

$3 (\cos (π) + i \sin (π))$

Do you know how to Convert to Trigonometric Form 3(cos(pi)+isin(pi))? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.