The general formula to figure out the modulus is

$\displaystyle \left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\displaystyle \left| ( 8 + 5 i ) ( -10 + 3 i ) \right| = \sqrt{ 64 + 25 }\cdot \sqrt{ 100 + 9 }$

$\displaystyle = \sqrt{89} \cdot \sqrt{109}$

$\displaystyle = \sqrt{9701}$

The general formula to figure out the modulus is

$\displaystyle \left|a+bi\right|=\sqrt{a^2+b^2}$.

We apply this notion to get.

$\displaystyle \left| 9 - 7 i \right| = \sqrt{ 81 + 49 }$

$\displaystyle = \sqrt{130}$

Converting from rectangular to polar coordinates gives us

$\displaystyle r = \sqrt{x^2+y^2} = \sqrt{3+1} = 2$

$\displaystyle \theta = tan^{-1}(y/x) = tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$

So 

$\displaystyle (\sqrt{3}-i)^6 = (2e^{-\frac{\pi}{6}i})^6 = 64e^{-\pi i} = -64$

Converting from Rectangular to Polar Coordinates

$\displaystyle (-16)^\frac{1}{4} = (16e^{i\pi + 2\pi k})^\frac{1}{4} = 2e^{i\frac{\pi}{4} + \frac{\pi}{2}k}$

Evaluating for $\displaystyle k = 0,1,2,3$

we get that 

$\displaystyle \pm\sqrt{2}(1+i) \text{ and } \pm\sqrt{2}(1-i)$

A 5th root of unity is a complex number $\displaystyle z$ such that $\displaystyle z^5=1$. Manipulating this equation yields $\displaystyle z^5-1=0$.

The magnitude of a complex number $\displaystyle a+bi$ is defined as

$\displaystyle \vert a+bi\vert=\sqrt{a^2+b^2}.$

So the modulus of $\displaystyle 5+2i$ is

$\displaystyle \vert5+2i\vert=\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$.

The magnitude of a complex number $\displaystyle a+bi$ is defined as

$\displaystyle \vert a+bi\vert=\sqrt{a^2+b^2}.$

So the modulus of $\displaystyle 3-i$ is

$\displaystyle \vert3-i\vert=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$.

The magnitude of a complex number $\displaystyle a+bi$ is defined as

$\displaystyle \vert a+bi\vert=\sqrt{a^2+b^2}.$

Because the complex number $\displaystyle 8$ has no imaginary part, we can write it in the form $\displaystyle 8+0i$. Then the modulus of $\displaystyle 8+0i$ is

$\displaystyle \vert8+0i\vert=\sqrt{8^2+0^2}=\sqrt{64+0}=\sqrt{64}=8$.

Note that the complex number $\displaystyle 1+i$ lies in the first quadrant of the complex plane.

For a complex number $\displaystyle z=a+bi$, the argument of $\displaystyle z$ is defined as the real number $\displaystyle \theta$ such that

$\displaystyle \tan(\theta)=\frac{b}{a}$,

where $\displaystyle -\pi< \theta\leq\pi$ is in radians.

Then the argument of $\displaystyle 1+i$ is

$\displaystyle \tan(\theta)=\frac{1}{1}=1$

$\displaystyle \theta=\frac{-3\pi}{4},\frac{\pi}{4}$.

The angle $\displaystyle \frac{-3\pi}{4}$ lies in the third quadrant of the complex plane, but the angle $\displaystyle \frac{\pi}{4}$ lies in the first quadrant, as does $\displaystyle 1+i$. So $\displaystyle \theta=\frac{\pi}{4}$.

Note that the complex number $\displaystyle 2-3i$ lies in the fourth quadrant of the complex plane.

For a complex number $\displaystyle z=a+bi$, the argument of $\displaystyle z$ is defined as the real number $\displaystyle \theta$ such that

$\displaystyle \tan(\theta)=\frac{b}{a}$,

where $\displaystyle -\pi< \theta\leq\pi$ is in radians.

Then the argument of $\displaystyle 2-3i$ is

$\displaystyle \tan(\theta)=\frac{-3}{2}$

$\displaystyle \theta=-0.98,2.16$.

The angle $\displaystyle 2.16$ lies in the second quadrant of the complex plane, but the angle $\displaystyle -0.98$ lies in the fourth quadrant, as does $\displaystyle 2-3i$. So $\displaystyle \theta=-0.98$.