Answer:
The fifth root is 2[cos(56°) + i sin(56°)]
Explanation:
* To solve this problem we must revise De Moiver's rule
- In the complex number with polar form
∵ z = r(cosФ + i sinФ)
∴ z^n = r^n(cos(nФ) + i sin(nФ))
* In the problem
- The fifth root means z^(1/5)
- We can put 32 as a form a^n
∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^5
∴ z = 2^5[cos(280°) + i sin(280°)]
* Lets find z^(1/5)
![*z^{(1)/(5)}=[2^(5)]^{(1)/(5) } (cos((1)/(5))(280)+isin((1)/(5))(280)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/21zzu3483tcoamdc71eksvdao1cimiei2x.png)

∴ z^(1/5) = 2[cos(56) + i sin(56)]
* The fifth root of 32[cos(280°) + i sin(280°)] is 2[cos(56°) + i sin(56°)]