Respuesta :
z1 × z2 = 32(cos(40° + 135°) + i sin ( 40° + 135°)
z1 × z2 = 32 (cos(175°) +i sin (175°)
z1 × z2 = 32 (cos(175°) +i sin (175°)
Answer:
Option D is correct.i.e., [tex]32\times(cos175^{\circ}+\imath sin175^{\circ})[/tex]
Step-by-step explanation:
Given: [tex]z_1=8(cos40^{\circ}+\imath sin40^{\circ})[/tex] and [tex]z_2=4(cos135^{\circ}+\imath sin135^{\cric})[/tex]
To find: Product of [tex]z_1[/tex] and [tex]z_2[/tex]
[tex]z_1\times z_2[/tex]
[tex]\implies8(cos40^{\circ}+\imath sin40^{\circ})\times4(cos135^{\circ}+\imath sin135^{\circ})[/tex]
[tex]\implies8\times4\times(cos40^{\circ}+\imath sin40^{\circ})(cos135^{\circ}+\imath sin135^{\circ})[/tex]
[tex]\implies32\times(cos40^{\circ}\times cos135^{\circ}+ cos40^{\circ}\times\imath sin135^{\circ}+\imath sin40^{\circ}\times cos135^{\circ}+\imath sin40^{\circ}\times\imath sin135^{\circ})[/tex]
[tex]\implies32\times(cos40^{\circ}\times cos135^{\circ}+\imath cos40^{\circ}\times sin135^{\circ}+\imath sin40^{\circ}\times cos135^{\circ}+(\imath)^2 sin40^{\circ}\times sin135^{\circ})[/tex]
[tex]\implies32\times(cos40^{\circ}\times cos135^{\circ}+\imath cos40^{\circ}\times sin135^{\circ}+\imath sin40^{\circ}\times cos135^{\circ}-1\times sin40^{\circ}\times sin135^{\circ})[/tex]
[tex]\implies32\times(cos40^{\circ}\times cos135^{\circ}-sin40^{\circ}\times sin135^{\circ}+\imath (cos40^{\circ}\times sin135^{\circ}+sin40^{\circ}\times cos135^{\circ}))[/tex]
Using, Cos( A + B ) = CosA CosB - SinA SinB
and Sin( A + B ) = SinA CosB + CosA SinB
[tex]\implies32\times(cos(40^{\circ}+135^{\cric})+\imath (sin(40^{\cric}+135^{\circ}))[/tex]
[tex]\implies32\times(cos175{\circ}+\imath sin175^{\cric})[/tex]
Therefore, Option D is correct.i.e., [tex]32\times(cos175^{\circ}+\imath sin175^{\circ})[/tex]
