Answer:
a) The standard form of [tex]z = 4\cdot (\cos 60^{\circ}+i\cdot \sin 60^{\circ})[/tex] is [tex]z = 2 + i\cdot 2\sqrt{3}[/tex], b) [tex]z = (2+i\cdot 3)^{6} = 1219.585 + i \cdot 1829.381[/tex].
Step-by-step explanation:
a) The standard form of the complex number is [tex]z = a + i\cdot b[/tex], [tex]\forall \,a,b \in \mathbb{R}[/tex]. If we get that [tex]z = 4\cdot (\cos 60^{\circ}+i\cdot \sin 60^{\circ})[/tex], whose standard form is obtained by algebraic means:
1) [tex]z = 4\cdot (\cos 60^{\circ}+i\cdot \sin 60^{\circ})[/tex] Given
2) [tex]z = (4\cdot \cos 60^{\circ})+i\cdot (4\cdot \sin 60^{\circ})[/tex] Distributive and Associative properties.
3) [tex]z = 2 + i\cdot 2\sqrt{3}[/tex] Multiplication/Result.
The standard form of [tex]z = 4\cdot (\cos 60^{\circ}+i\cdot \sin 60^{\circ})[/tex] is [tex]z = 2 + i\cdot 2\sqrt{3}[/tex].
b) The De Moivre's Theorem states that:
[tex]z = (a+i\cdot b)^{n}= r^{n}\cdot (\cos \theta + i\cdot \sin \theta)[/tex]
Where:
[tex]r =\sqrt{a^{2}+b^{2}}[/tex] and [tex]\theta = \tan^{-1} \left(\frac{b}{a}\right)[/tex].
If we know that [tex]z = (2+i\cdot 3)^{6}[/tex], then:
[tex]r = \sqrt{2^{2}+3^{2}}[/tex]
[tex]r =\sqrt{13}[/tex]
[tex]r \approx 3.606[/tex]
[tex]\theta = \tan^{-1}\left(\frac{3}{2} \right)[/tex]
[tex]\theta \approx 56.310^{\circ}[/tex]
The resulting expression is:
[tex]z = 3.606^{6}\cdot (\cos 56.310^{\circ}+i\cdot \sin 56.310^{\circ})[/tex]
[tex]z = 1219.585+i\cdot 1829.381[/tex]
Therefore, [tex]z = (2+i\cdot 3)^{6} = 1219.585 + i \cdot 1829.381[/tex].
