Given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle x$-coordinate is  $\displaystyle x= r \cos \theta$. We can find this coordinate by substituting $\displaystyle r = 2.1, \theta = \frac{7\pi }{3}$:

$\displaystyle x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $\displaystyle (r,\theta )$, the  $\displaystyle y$-coordinate is  $\displaystyle y= r \sin \theta$.  We can find this coordinate by substituting $\displaystyle r = 2.1, \theta = \frac{7\pi }{3}$:

$\displaystyle y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\displaystyle \left(2.1, \frac{7\pi }{3}\right )$ are $\displaystyle (1.05, 1.82)$.

To convert this number to rectangular form, first think about what $\displaystyle \cos240^\circ$ and $\displaystyle \sin240^\circ$ are equal to. Because $\displaystyle 240^\circ=180^\circ+60^\circ$, we can use a 30-60-90^o reference triangle in the 3rd quadrant to determine these values. 

$\displaystyle \cos240^\circ=-\frac{1}{2}$

$\displaystyle \sin240^\circ=-\frac{\sqrt{3}}{2}$

Now plug these in and continue solving:

$\displaystyle z=4(\cos240^\circ+i\sin240^\circ)$

$\displaystyle z=4(-\frac{1}{2}-i\frac{\sqrt{3}}{2})$

$\displaystyle z=-2-2\sqrt{3}i$

This problem has given us formulas, so we just need to plug in $\displaystyle x=7$ and $\displaystyle y=13$ and solve. 

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r=\sqrt{7^2+13^2}$

$\displaystyle r=\sqrt{218}$

$\displaystyle \theta=\tan^-^1\left (\frac{y}{x} \right )$

$\displaystyle \theta=\tan^-^1\left (\frac{13}{7} \right )$

$\displaystyle \theta=61.7^\circ$

$\displaystyle z=r(\cos\theta+i\sin\theta)$

$\displaystyle z=\sqrt{218}(\cos61.7^\circ+i\sin61.7^\circ)$

To convert this number to rectangular form, first think about what $\displaystyle \cos30^\circ$and $\displaystyle \sin30^\circ$ are equal to. We can use a 30-60-90^o reference triangle in the 1st quadrant to determine these values. 

$\displaystyle \cos30^\circ=\frac{\sqrt{3}}{2}$

$\displaystyle \sin30^\circ=\frac{1}{2}$

Next, plug these values in and simplify:

$\displaystyle z=3(\cos30^\circ+i\sin30^\circ)$

$\displaystyle z=3(\frac{\sqrt{3}}{2}+\frac{1}{2}i)$

$\displaystyle z=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$

This problem has given us formulas, so we just need to plug in $\displaystyle x=2$ and $\displaystyle y=-5$ and solve. 

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r=\sqrt{2^2+(-5)^2}$

$\displaystyle r=\sqrt{29}$

$\displaystyle \theta=\tan^-^1\left (\frac{y}{x} \right )$

$\displaystyle \theta=\tan^-^1\left (\frac{-5}{2} \right )$

$\displaystyle \theta=-68.2^\circ$

$\displaystyle z=r(\cos\theta+i\sin\theta)$

$\displaystyle z=\sqrt{29}(\cos(-68.2^\circ)+i\sin(-68.2^\circ))$

In order to complete this problem, you must understand three formulas that allow you to convert from the rectangular form of a complex number to the polar form of a complex number. These formulas are $\displaystyle r=\sqrt{x^2+y^2}$,  $\displaystyle \theta=\tan^-^1\left (\frac{y}{x} \right )$, and the polar form $\displaystyle z=r(\cos\theta+i\sin\theta)$. Additionally, understand that based on the given info, $\displaystyle x=-1$ and $\displaystyle y=-4$. Begin by finding the modulus:

$\displaystyle r=\sqrt{x^2+y^2}$

$\displaystyle r=\sqrt{(-1)^2+(-4)^2}$

$\displaystyle r=\sqrt{17}$

Next, let's find the angle $\displaystyle \theta$, also referred to as the amplitude of the complex number. 

$\displaystyle \theta=\tan^-^1\left (\frac{y}{x} \right )$

$\displaystyle \theta=\tan^-^1\left (\frac{-1}{-4} \right )$

$\displaystyle \theta=\tan^-^1\left (\frac{1}{4} \right )$

$\displaystyle \theta=14.04^\circ$

Finally, plug each of these into the polar form of a complex number: 

$\displaystyle z=r(\cos\theta+i\sin\theta)$

$\displaystyle z=\sqrt{17}(\cos14.04^\circ+i\sin14.04^\circ)$

The modulus of the product of two complex numbers is the product of their moduli, and the amplitude of the product is the sum of their amplitudes. 

Therefore, the new modulus will be $\displaystyle 2\cdot 3=6$ and the new amplitude will be $\displaystyle 70^\circ+20^\circ=90^\circ$. Therefore

$\displaystyle 2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$$\displaystyle =6(\cos90^\circ+i\sin90^\circ)$

We must also express this in rectangular form, which we can do by substituting $\displaystyle \cos90^\circ=0$ and $\displaystyle \sin90^\circ=1$. We get:

$\displaystyle 6(\cos90^\circ+i\sin90^\circ)$

$\displaystyle =6(0+1i)$

$\displaystyle =6i$

The modulus of the quotient of two complex numbers is the modulus of the dividend divided by the modulus of the divisor. The amplitude of the quotient is the amplitude of the dividend minus the amplitude of hte divisor.

(a) The modulus for $\displaystyle \frac{z_1}{z_2}$ is equal to $\displaystyle \frac{10}{2}=5$. The amplitude for $\displaystyle \frac{z_1}{z_2}$ is equal to $\displaystyle 40^\circ-220^\circ=-180^\circ=180^\circ$. (We have chosen to represent this as the coterminal angle $\displaystyle 180^\circ$ rather than $\displaystyle -180^\circ$ as it is more conventional to represent angle measures as a positive angle between $\displaystyle 0^\circ$ and $\displaystyle 360^\circ$.) Putting this together, we get $\displaystyle \frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$. To represent this in rectangular form, substitute $\displaystyle \cos180^\circ=-1$ and $\displaystyle \sin180^\circ=0$ to get $\displaystyle \frac{z_1}{z_2}=5(-1+0i)=-5$.

(b) The modulus for $\displaystyle \frac{z_2}{z_1}$ is equal to $\displaystyle \frac{2}{10}=\frac{1}{5}$. The amplitude for $\displaystyle \frac{z_2}{z_1}$ is equal to $\displaystyle 220^\circ-40^\circ=180^\circ$. Putting this together, we get $\displaystyle \frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$. To represent this in rectangular form, substitute $\displaystyle \cos180^\circ=-1$ and $\displaystyle \sin180^\circ=0$ to get $\displaystyle \frac{z_2}{z_1}=\frac{1}{5}(-1+0i)=-\frac{1}{5}$.