From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction.

Figure 1

Using the formula, we have

[latex]\begin{align}&|z|=\sqrt{{x}^{2}+{y}^{2}}\\ &|z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}} \\ &|z|=\sqrt{5+1} \\ &|z|=\sqrt{6} \end{align}[/latex]

Figure 3

Using the formula, we have

[latex]\begin{align}&|z|=\sqrt{{x}^{2}+{y}^{2}} \\ &|z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}} \\ &|z|=\sqrt{9+16} \\ &|z|=\sqrt{25}\\ &|z|=5 \end{align}[/latex]

The absolute value [latex]z[/latex] is 5.

Figure 4

On the complex plane, the number [latex]z=4i[/latex] is the same as [latex]z=0+4i[/latex]. Writing it in polar form, we have to calculate [latex]r[/latex] first.

[latex]\begin{align}&r=\sqrt{{x}^{2}+{y}^{2}} \\ &r=\sqrt{{0}^{2}+{4}^{2}} \\ &r=\sqrt{16} \\ &r=4 \end{align}[/latex]

Next, we look at [latex]x[/latex]. If [latex]x=r\cos \theta[/latex], and [latex]x=0[/latex], then [latex]\theta =\frac{\pi }{2}[/latex]. In polar coordinates, the complex number [latex]z=0+4i[/latex] can be written as [latex]z=4\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)[/latex] or [latex]4\text{cis}\left(\frac{\pi }{2}\right)[/latex].

Figure 6

First, find the value of [latex]r[/latex].

[latex]\begin{align}&r=\sqrt{{x}^{2}+{y}^{2}} \\ &r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)} \\ &r=\sqrt{32} \\ &r=4\sqrt{2} \end{align}[/latex]

Find the angle [latex]\theta[/latex] using the formula:

[latex]\begin{align}&\cos \theta =\frac{x}{r} \\ &\cos \theta =\frac{-4}{4\sqrt{2}} \\ &\cos \theta =-\frac{1}{\sqrt{2}} \\ &\theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4} \end{align}[/latex]

Thus, the solution is [latex]4\sqrt{2}\cos\left(\frac{3\pi }{4}\right)[/latex].

We begin by evaluating the trigonometric expressions.

[latex]\begin{gathered}\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}\end{gathered}[/latex]

After substitution, the complex number is

[latex]z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)[/latex]

We apply the distributive property:

[latex]\begin{align}z&=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right) \\ &=\left(12\right)\frac{\sqrt{3}}{2}+\left(12\right)\frac{1}{2}i \\ &=6\sqrt{3}+6i \end{align}[/latex]

The rectangular form of the given point in complex form is [latex]6\sqrt{3}+6i[/latex].

If [latex]\tan \theta =\frac{5}{12}[/latex], and [latex]\tan \theta =\frac{y}{x}[/latex], we first determine [latex]r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.}[/latex] We then find [latex]\cos \theta =\frac{x}{r}[/latex] and [latex]\sin \theta =\frac{y}{r}[/latex].

[latex]\begin{align}z&=13\left(\cos \theta +i\sin \theta \right) \\ &=13\left(\frac{12}{13}+\frac{5}{13}i\right) \\ &=12+5i \end{align}[/latex]

The rectangular form of the given number in complex form is [latex]12+5i[/latex].

Follow the formula

[latex]\begin{align}&{z}_{1}{z}_{2}=4\cdot 2\left[\cos \left(80^\circ +145^\circ \right)+i\sin \left(80^\circ +145^\circ \right)\right] \\ &{z}_{1}{z}_{2}=8\left[\cos \left(225^\circ \right)+i\sin \left(225^\circ \right)\right] \\ &{z}_{1}{z}_{2}=8\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right] \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right] \\ &{z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2} \end{align}[/latex]

Using the formula, we have

[latex]\begin{align}&\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\cos \left(213^\circ -33^\circ \right)+i\sin \left(213^\circ -33^\circ \right)\right] \\ &\frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\cos \left(180^\circ \right)+i\sin \left(180^\circ \right)\right] \\ &\frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right] \\ &\frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i \\ &\frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2} \end{align}[/latex]

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write [latex]\left(1+i\right)[/latex] in polar form. Let us find [latex]r[/latex].

[latex]\begin{align}&r=\sqrt{{x}^{2}+{y}^{2}} \\ &r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}} \\ &r=\sqrt{2} \end{align}[/latex]

Then we find [latex]\theta[/latex]. Using the formula [latex]\tan \theta =\frac{y}{x}[/latex] gives

[latex]\begin{align}&\tan \theta =\frac{1}{1} \\ &\tan \theta =1 \\ &\theta =\frac{\pi }{4} \end{align}[/latex]

Use De Moivre’s Theorem to evaluate the expression.

[latex]\begin{align}&{\left(a+bi\right)}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\\ &{\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\cos \left(5\cdot \frac{\pi }{4}\right)+i\sin \left(5\cdot \frac{\pi }{4}\right)\right] \\ &{\left(1+i\right)}^{5}=4\sqrt{2}\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right] \\ &{\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right] \\ &{\left(1+i\right)}^{5}=-4 - 4i \end{align}[/latex]

We have

[latex]\begin{align}&{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left[\cos \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)\right] \\ &{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)\right] \end{align}[/latex]

There will be three roots: [latex]k=0,1,2[/latex]. When [latex]k=0[/latex], we have

[latex]{z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)[/latex]

When [latex]k=1[/latex], we have

[latex]\begin{align}&{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)\right] && \text{ Add }\frac{2\left(1\right)\pi }{3}\text{ to each angle.} \\ &{z}^{\frac{1}{3}}=2\left(\cos \left(\frac{8\pi }{9}\right)+i\sin \left(\frac{8\pi }{9}\right)\right) \end{align}[/latex]

When [latex]k=2[/latex], we have

[latex]\begin{align}&{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)\right]&& \text{Add }\frac{2\left(2\right)\pi }{3}\text{ to each angle.} \\ &{z}^{\frac{1}{3}}=2\left(\cos \left(\frac{14\pi }{9}\right)+i\sin \left(\frac{14\pi }{9}\right)\right)\end{align}[/latex]

Remember to find the common denominator to simplify fractions in situations like this one. For [latex]k=1[/latex], the angle simplification is

[latex]\begin{align}\frac{\frac{2\pi }{3}}{3}+\frac{2\left(1\right)\pi }{3}&=\frac{2\pi }{3}\left(\frac{1}{3}\right)+\frac{2\left(1\right)\pi }{3}\left(\frac{3}{3}\right)\\ &=\frac{2\pi }{9}+\frac{6\pi }{9} \\ &=\frac{8\pi }{9} \end{align}[/latex]