Answer:
A
Explanation:
Begin with Euler's formula. I am assuming youre aware of this if youre taking complex algebra? You can prove Euler's formula by doing a Maclaurin expansion of cosine, sine, and e^x. Euler's formula states that:
![e^(ix)=cos(x)+isin(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/6w9jyx3eqpm74ubduu86a7oblaxcrkib57.png)
We can put the first complex number in exponetial form by noticing the input is 2pi/3. The second complex number has an input of pi/3. Therefore:
![z_1=8e^(i(2\pi /3))](https://img.qammunity.org/2023/formulas/mathematics/high-school/7dxdjrn2jtb3wgp1ur3gh29l5guw0o4izm.png)
![z_2=0.5e^(i(\pi /3))](https://img.qammunity.org/2023/formulas/mathematics/high-school/h2r477ep5idt32gqwk3y0t7979v42d0tfl.png)
Then:
![(z_1)/(z_2) =(8e^(i(2\pi /3)))/(0.5e^(i(\pi /3)))](https://img.qammunity.org/2023/formulas/mathematics/high-school/jlol8mmma4jv7g4zkjgp737kw9qp1p7ytj.png)
Simplify the coefficients to get:
![(z_1)/(z_2) =(16e^(i(2\pi /3)))/(e^(i(\pi /3)))](https://img.qammunity.org/2023/formulas/mathematics/high-school/uro53jqknxdpnh4cxy19fofdvujhq0fat6.png)
When you divide exponentials, you subtract the exponents. Therefore:
![(z_1)/(z_2) =16e^(i(2\pi /3)-i(\pi /3))=16e^{i(\pi /3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/blt4azr2ni72u2a8crmz2c3mcjtrc934fa.png)
Put it back into trigonemtric form using Euler's formula:
![16e^(i(\pi /3))=16cos(\pi /3)+i16sin(\pi /3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/gu2r8hx9ysxag3jjnmvksd7qr1fwix20n1.png)
Cosine of pi/3 is 0.5, and sine of pi/3 is square root of 3 over 2. We have:
![16e^(i(\pi /3))=16cos(\pi /3)+i16sin(\pi /3)=16*0.5+16*(√(3) )/(2) *i=8+8√(3) i](https://img.qammunity.org/2023/formulas/mathematics/high-school/lu43dkt4jxr5rcghi2fr56hoovfzxe7l5r.png)