Question

Write the complex number 4(cos 60 + i sin 60) in standard form 10. Use DeMoivre's Theorem to find (2+3i)6

Answer 1

a) The standard form of
`z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))` is
`z = 2 + i\cdot 2√(3)`, b)
`z = (2+i\cdot 3)^(6) = 1219.585 + i \cdot 1829.381`.

Explanation:

a) The standard form of the complex number is
`z = a + i\cdot b`,
`\forall \,a,b \in \mathbb{R}`. If we get that
`z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))`, whose standard form is obtained by algebraic means:

1)
`z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))` Given

2)
`z = (4\cdot \cos 60^(\circ))+i\cdot (4\cdot \sin 60^(\circ))` Distributive and Associative properties.

3)
`z = 2 + i\cdot 2√(3)` Multiplication/Result.

The standard form of
`z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))` is
`z = 2 + i\cdot 2√(3)`.

b) The De Moivre's Theorem states that:

`z = (a+i\cdot b)^(n)= r^(n)\cdot (\cos \theta + i\cdot \sin \theta)`

Where:

`r =\sqrt{a^(2)+b^(2)}` and
`\theta = \tan^(-1) \left((b)/(a)\right)`.

If we know that
`z = (2+i\cdot 3)^(6)`, then:

`r = \sqrt{2^(2)+3^(2)}`

`r =√(13)`

`r \approx 3.606`

`\theta = \tan^(-1)\left((3)/(2) \right)`

`\theta \approx 56.310^(\circ)`

The resulting expression is:

`z = 3.606^(6)\cdot (\cos 56.310^(\circ)+i\cdot \sin 56.310^(\circ))`

`z = 1219.585+i\cdot 1829.381`

Therefore,
`z = (2+i\cdot 3)^(6) = 1219.585 + i \cdot 1829.381`.