Calculus Examples

$| \frac{3 π}{4} |$

Step 1

$\frac{3 π}{4}$ is approximately $2.35619449$ which is positive so remove the absolute value

$\frac{3 π}{4}$

Step 2

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 3

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 4

Substitute the actual values of $a = \frac{3 π}{4}$ and $b = 0$ .

$| z | = \sqrt{0^{2} + {(\frac{3 π}{4})}^{2}}$

Step 5

Find $| z |$ .

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Step 5.1

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

$| z | = \sqrt{0 + {(\frac{3 π}{4})}^{2}}$

Step 5.2

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 5.2.1

Apply the product rule to $\frac{3 π}{4}$ .

$| z | = \sqrt{0 + \frac{{(3 π)}^{2}}{4^{2}}}$

Step 5.2.2

Apply the product rule to $3 π$ .

$| z | = \sqrt{0 + \frac{3^{2} π^{2}}{4^{2}}}$

Step 5.3

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

$| z | = \sqrt{0 + \frac{9 π^{2}}{4^{2}}}$

Step 5.4

Raise $4$ to the power of $2$ .

$| z | = \sqrt{0 + \frac{9 π^{2}}{16}}$

Step 5.5

Add $0$ and $\frac{9 π^{2}}{16}$ .

$| z | = \sqrt{\frac{9 π^{2}}{16}}$

Step 5.6

Rewrite $\sqrt{\frac{9 π^{2}}{16}}$ as $\frac{\sqrt{9 π^{2}}}{\sqrt{16}}$ .

$| z | = \frac{\sqrt{9 π^{2}}}{\sqrt{16}}$

Step 5.7

Simplify the numerator.

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Step 5.7.1

Rewrite $9 π^{2}$ as ${(3 π)}^{2}$ .

$| z | = \frac{\sqrt{{(3 π)}^{2}}}{\sqrt{16}}$

Step 5.7.2

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

$| z | = \frac{3 π}{\sqrt{16}}$

Step 5.8

Simplify the denominator.

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Step 5.8.1

Rewrite $16$ as $4^{2}$ .

$| z | = \frac{3 π}{\sqrt{4^{2}}}$

Step 5.8.2

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

$| z | = \frac{3 π}{4}$

Step 6

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

$θ = \arctan (\frac{0}{\frac{3 π}{4}})$

Step 7

Substitute the values of $θ = \arctan (\frac{0}{\frac{3 π}{4}})$ and $| z | = \frac{3 π}{4}$ .

$\frac{3 π}{4 (\cos (\arctan (\frac{0}{\frac{3 π}{4}})) + i \sin (\arctan (\frac{0}{\frac{3 π}{4}})))}$