Question

10 POINTS!

Find the product of z1 and z2, where z1 = 8(cos 40° + i sin 40°) and z2 = 4(cos 135° + i sin 135°).

A) 32(cos 40° + i sin 40°)

B) 12(cos 175° + i sin 175°)

C) 32(cos 5400° + i sin 5400°)

D) 32(cos 175° + i sin 175°)

Answer 1

z1 × z2 = 32(cos(40° + 135°) + i sin ( 40° + 135°)
z1 × z2 = 32 (cos(175°) +i sin (175°)

Answer 2

Option D is correct.i.e.,
`32*(cos175^(\circ)+\imath sin175^(\circ))`

Explanation:

Given:
`z_1=8(cos40^(\circ)+\imath sin40^(\circ))` and
`z_2=4(cos135^(\circ)+\imath sin135^(\cric))`

To find: Product of
`z_1` and
`z_2`

`z_1* z_2`

`\implies8(cos40^(\circ)+\imath sin40^(\circ))*4(cos135^(\circ)+\imath sin135^(\circ))`

`\implies8*4*(cos40^(\circ)+\imath sin40^(\circ))(cos135^(\circ)+\imath sin135^(\circ))`

`\implies32*(cos40^(\circ)* cos135^(\circ)+ cos40^(\circ)*\imath sin135^(\circ)+\imath sin40^(\circ)* cos135^(\circ)+\imath sin40^(\circ)*\imath sin135^(\circ))`

`\implies32*(cos40^(\circ)* cos135^(\circ)+\imath cos40^(\circ)* sin135^(\circ)+\imath sin40^(\circ)* cos135^(\circ)+(\imath)^2 sin40^(\circ)* sin135^(\circ))`

`\implies32*(cos40^(\circ)* cos135^(\circ)+\imath cos40^(\circ)* sin135^(\circ)+\imath sin40^(\circ)* cos135^(\circ)-1* sin40^(\circ)* sin135^(\circ))`

`\implies32*(cos40^(\circ)* cos135^(\circ)-sin40^(\circ)* sin135^(\circ)+\imath (cos40^(\circ)* sin135^(\circ)+sin40^(\circ)* cos135^(\circ)))`

Using, Cos( A + B ) = CosA CosB - SinA SinB

and Sin( A + B ) = SinA CosB + CosA SinB

`\implies32*(cos(40^(\circ)+135^(\cric))+\imath (sin(40^(\cric)+135^(\circ)))`

`\implies32*(cos175{\circ}+\imath sin175^(\cric))`

Therefore, Option D is correct.i.e.,
`32*(cos175^(\circ)+\imath sin175^(\circ))`