Question

04 - Considere o número complexo z =
$\frac{1}{2} ( \sqrt{3} + i)$
a) Calcule o módulo de z e escreva a forma polar de z.


b) calcule o valor da expressão
${z}^{27} + {z}^{24} + 1$
​

Answer 1

Resposta:

a)

|z| = 1

z=cosπ/6+i.senπ/6

b)

$\displaystyle z^{27}+z^{24}+1=2+i$

Explicação passo-a-passo:

z=1/2(√3+i)=√3/2+i/2

a)

Módulo de z:

|z| = √[(√3/2)²+(1/2)²]=√[(3/4)+(1/4)]=√1=1

Argumento:

cosΘ=√3/2÷1=√3/2

senΘ=1/2÷1=1/2

∴ Θ=π/6 rad

z=1(cosπ/6+i.senπ/6)=cosπ/6+i.senπ/6

b)

Fórmula de Moivre

$\displaystyle z^n=|z|^n[cos(n.\theta)+i.sen(n.\theta)]$

$\displaystyle z^{27}=(1) ^{27}[cos(27.\frac{\pi}{6} )+i.sen(27.\frac{\pi}{6})]=cos(9.\frac{\pi}{2} )+i.sen(9.\frac{\pi}{2})$

cos(9π/2)=cos(π/2)=0

sen(9π/2)=sen(π/2)=1

$\displaystyle z^{27}=cos(9.\frac{\pi}{2} )+i.sen(9.\frac{\pi}{2})=0+i.1=i$

$\displaystyle z^{24}=1^{24}[cos(24.\frac{\pi}{6} )+i.sen(24.\frac{\pi}{6})]=cos(4\pi)+i.sen(4\pi)$

cos(4π)=cos(0)=1

sen(4π)=sen(0)=0

$\displaystyle z^{24}=1+i.0=1$

$\displaystyle z^{27}+z^{24}+1=i+1+1=2+i$