Question

POR FAVOR ME AJUDA
Dado o número complexo Z=4(cos 240° + i sen240°), determine z^‐2.

Answer 1

Resposta:

cos(240)=cos(180+60)=cos(180)*cos(60)-sen(180)*sen(60)

= -1 * 1/2 - 0 * sen(60)= -1/2

sen(240)=sen(180+60) =sen(180) *cos(60) + sen(60)* cos(180)

=0 * cos (60) +√3/2 *(-1)= -√3/2

Z=4(cos 240° + i sen240°),

Z=4(-1/2 + i *( -√3/2))

Z= -2 - 2√3 i

Z⁻² =1/Z²

= 1/(-2 - 2√3 i)²

=1/(4+8√3 i+(2√3)²*i²)

=1/(4+8√3i+(12)*(-1))

=1/(4+8√3i-12)

=1/(8√3i-8)

= (-1/8)  * 1/(1-√3i)

= (-1/8)  * (1+√3i)/(1-√3i)(1+√3i)

= (-1/8)  * (1+√3i)/(1²-√3² *i²)

= (-1/8)  * (1+√3i)/(1-3 *(-1))

= (-1/8)  * (1+√3i)/4

= -1/32 - (i√3) /32