Answer:
Attachment 1 : Option C
Attachment 2 : Option A
Explanation:
( 1 ) Expressing the product of z1 and z2 would be as follows,
![14\left[\cos \left((\pi \:)/(5)\right)+i\sin \left((\pi \:\:)/(5)\right)\right]\cdot \:2√(2)\left[\cos \left((3\pi \:)/(2)\right)+i\sin \left((3\pi \:\:)/(2)\right)\right]](https://img.qammunity.org/2021/formulas/mathematics/college/rj9jx84wzuquke5fmdhoeo66ckhhg02o1n.png)
Now to solve such problems, you will need to know what cos(π / 5) is, sin(π / 5) etc. If you don't know their exact value, I would recommend you use a calculator,
cos(π / 5) =
,
sin(π / 5) =

cos(3π / 2) = 0,
sin(3π / 2) = - 1
Let's substitute those values in our expression,
And now simplify the expression,

The exact value of
=
and
=
Therefore we have the expression
, which is close to option c. As you can see they approximated the solution.
( 2 ) Here we will apply the following trivial identities,
cos(π / 3) =
,
sin(π / 3) =
,
cos(- π / 6) =
,
sin(- π / 6) =

Substitute into the following expression, representing the quotient of the given values of z1 and z2,
⇒
![15\left[(1)/(2)+(√(3))/(2)\right]/ \:3√(2)\left[(√(3))/(2)+-(1)/(2)\right]](https://img.qammunity.org/2021/formulas/mathematics/college/rccnglcfhp0rqq7qipf1f8jfhi793xyp9g.png)
The simplified expression will be the following,
or in other words
or

The solution will be option a, as you can see.