The correct answer is 

$\displaystyle \small 5e^{\arctan\frac{4}{3}i}$

The polar form of a complex number $\displaystyle \small a+bi$ is $\displaystyle \small re^{i\theta}$ where $\displaystyle \small r$ is the modulus of the complex number and $\displaystyle \small \theta$ is the angle in radians between the real axis and the line that passes through $\displaystyle \small a+bi$ ( $\displaystyle \small r=\sqrt{a^2+b^2}$  and $\displaystyle \small \theta = \arctan(b/a)$). We can solve for $\displaystyle \small r$ and $\displaystyle \small \theta$ easily for the complex number $\displaystyle \small 3+4i$:

$\displaystyle \small r = \sqrt{3^2+4^2}=\sqrt{25}=5$

$\displaystyle \small \theta = \arctan(4/3)$

which gives us

$\displaystyle \small 5e^{\arctan\frac{4}{3}i}$

Remember that the standard form of a complex number is: $\displaystyle z=a+bi$, which can be rewritten in polar form as: $\displaystyle z=r(cos(\theta)+isin(\theta))$.

To find r, we must find the length of the line $\displaystyle 4+3i$ by using the Pythagorean theorem:

$\displaystyle r=\sqrt{a^2+b^2}$

$\displaystyle r = \sqrt{4^2+3^2}$

    $\displaystyle =\sqrt{16+9}$

    $\displaystyle = \sqrt{25}$

    $\displaystyle =5$

To find $\displaystyle \theta$, we can use the equation 

$\displaystyle \theta=tan^{-1}\frac{b}{a}$

   $\displaystyle =tan^{-1}\frac{3}{4}$

   $\displaystyle \approx 0.64$

Note that this value is in radians, NOT degrees.

Thus, the polar form of this equation can be written as 

$\displaystyle z= 5(cos(0.64)+isin(0.64))$

$\displaystyle x^2 +y^2=r^2$

$\displaystyle x=r \cos\Theta$

$\displaystyle y= r \sin\Theta$

$\displaystyle \tan \Theta = \frac{y}{x}$

Given these identities, first solve for $\displaystyle r$ and $\displaystyle \Theta$. The polar form of a complex number is: $\displaystyle r\left [ \cos \Theta +i \sin \Theta \right ]$

$\displaystyle z=1+i$

$\displaystyle x=1$

$\displaystyle y=1$

$\displaystyle r=\sqrt{2}$

$\displaystyle \tan \Theta = 1$

$\displaystyle \tan \Theta = 1$ at $\displaystyle \frac{\pi}{4}$ (because the original point, (1,1) is in Quadrant 1)

Therefore...

$\displaystyle x=\sqrt{2} \cos(\frac{\pi}{4})=\sqrt{2} \:\frac{\sqrt{2}}{2}$

$\displaystyle y=\sqrt{2} \sin(\frac{\pi}{4})=\sqrt{2} \:\frac{\sqrt{2}}{2}$

$\displaystyle z=\sqrt{2}\left [\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right]$

First, find the radius $\displaystyle r$:

$\displaystyle r = \sqrt{4^2 + (-3)^2 } = \sqrt{16 + 9 } = \sqrt{25} = 5$

Then find the angle, thinking of the imaginary part as the height and the radius as the hypotenuse of a right triangle:

$\displaystyle \sin \theta = \frac{ -3}{5}$

$\displaystyle \theta = \sin ^{-1} (-\frac{3}{5}) \approx -36.870$ according to the calculator.

We can get the positive coterminal angle by adding $\displaystyle 360$:

$\displaystyle -36.87^o + 360^o = 323.13 ^o$

The polar form is

$\displaystyle 5 (\cos 323.13^o + i \sin 323.13^o)$

First find the radius, $\displaystyle r$:

$\displaystyle r = \sqrt{3^2 + 2^2 } = \sqrt{9 + 4 } =\sqrt{13}$

Now find the angle, thinking of the imaginary part as the height and the radius as the hypotenuse of a right triangle:

$\displaystyle \sin \theta = \frac{ 2 } {\sqrt{13 }}$

$\displaystyle \theta = \sin ^ {-1} (\frac{ 2 }{\sqrt{13}}) \approx 33.69$ according to the calculator.

This is an appropriate angle to stay with since this number should be in quadrant I.

The complex number in polar form is $\displaystyle \sqrt{13} (\cos 33.69^o + i \sin 33.69 ^o )$

First find $\displaystyle r$:

$\displaystyle r = \sqrt{(-2)^2 + 5^2 } = \sqrt{4 + 25 } = \sqrt{29}$

Now find the angle. Consider the imaginary part to be the height of a right triangle with hypotenuse $\displaystyle r$.

$\displaystyle \sin \theta = \frac{5}{\sqrt{29 } }$

$\displaystyle \theta = \sin ^{-1} (\frac{5}{\sqrt{29}})\approx 68.199 ^o$ according to the calculator.

What the calculator does not know is that this angle is actually located in quadrant II, since the real part is negative and the imaginary part is positive.

To find the angle in quadrant II whose sine is also $\displaystyle \frac{5 }{ \sqrt{29}}$, subtract from $\displaystyle 180$:

$\displaystyle 180^o - 68.199^o = 111.801^o$

The complex number in polar form is $\displaystyle \sqrt{29} (\cos 111.801^o + i \sin 111.801^o )$

To convert polar coordinates $\displaystyle (r, \theta)$ to rectangular coordinates $\displaystyle (x, y)$,

$\displaystyle x=r\cos\theta$

$\displaystyle y=r\sin\theta$

Using the information given in the question, 

$\displaystyle x=5\cos \frac{3\pi}{2}=5(0)=0$

$\displaystyle y=5 \sin \frac{3\pi}{2}=5(-1)=-5$

The rectangular coordinates are $\displaystyle (0, -5)$

This is the graph of a cardiod. Based on its orientation where the cusp [pointy part] is on the y-axis, it is a sine and not cosine function. The x-intercepts are at $\displaystyle \pm2$, so the first number must be 2. Since vertically the graph goes from 0 to 4, the second number must be 2, because $\displaystyle 2-2=0$ and $\displaystyle 2+2=4$.

This graph shows a rose curve with an odd number of petals.

This means that the equation will be in the form $\displaystyle r=asin(b\theta)$, where $\displaystyle a$ represents the length of the "petals" and $\displaystyle b$ represents the number of petals.

There are 5 petals of length 7.

This graph shows a rose curve with an even number of petals. The first petal also intersects with the x-axis. This means that the equation will be in the form $\displaystyle r=acos(b\theta)$ where $\displaystyle a$ is the length of each petal, and $\displaystyle b$ is half the number of petals. (Note that for an odd number of petals, the rose curve will have exactly $\displaystyle b$ petals). In this case, the petals have length 5, and there are 8 of them [half of 8 is 4].