Question

3 373,Consider the complex number z =+22What is 23?Hint: z has a modulus of 3 and an argument of 120°.Choose 1 answer:А-2727-13.5 +23.41-13.5 - 23.41

Answer 1

To answer this question, we can proceed as follows:

`z=-(3)/(2)+\frac{3\sqrt[]{3}}{2}i^{}\Rightarrow z^3=(-(3)/(2)+\frac{3\sqrt[]{3}}{2}i)^3`
`(-(3)/(2)+\frac{3\sqrt[]{3}i}{2})^3=(\frac{-3+3\sqrt[]{3}i}{2})^3=\frac{(-3+3\sqrt[]{3}i)^3}{2^3}`

We applied the exponent rule:

`((a)/(b))^c=(a^c)/(b^c)`

Then, we have:

`\frac{(-3+3\sqrt[]{3}i)^3}{2^3}=\frac{(-3+3\sqrt[]{3}i)^3}{8}`

Solving the numerator, we have:

`(a+b)^3=a^3+b^3+3ab(a+b)`

`(-3+3\sqrt[]{3}i)^3=(-3)^3+(3\sqrt[]{3}i)^3+3(-3)(3\sqrt[]{3}i)(-3+3\sqrt[]{3}i)`
`-27+81\sqrt[]{3}i^3-27\sqrt[]{3}i(-3+3\sqrt[]{3}i)`
`-27+81\sqrt[]{3}i^3+81\sqrt[]{3}i-27\cdot3\cdot(\sqrt[]{3})^2\cdot i^2`
`-27+81\sqrt[]{3}i^2\cdot i+81\sqrt[]{3}i-81\cdot3\cdot(-1)`
`-27+81\sqrt[]{3}(-1)\cdot i+81\sqrt[]{3}i+243`
`-27-81\sqrt[]{3}i+81\sqrt[]{3}i+243`
`-27+243=216`

Then, the numerator is equal to 216. The complete expression is:

`=\frac{(-3+3\sqrt[]{3}i)^3}{8}=(216)/(8)=27`

Therefore, we have that:

`z^3=(-(3)/(2)+\frac{3\sqrt[]{3}}{2}i)^3=27`

In summary, therefore, the value for z³ = 27 (option B).