Question

For questions 13-15, Let Z1=2(cos(pi/5)+i Sin(pi/5)) And Z2=8(cos(7pi/6)+i Sin(7pi/6)). Calculate The Following Keeping Your Answer In Polar Form. 13. z1z2 14. z2 15. z1/z2

Answer 1

13. 16(cos(41 π/30)+ isin(41 π/30))

14. Mine asked for z2 magnitude so I got 8 (magnitude is the same as modulus which is r)

15. 1/8 (cos(29 π/30)+ isin(29 π/30))

Explanation:

13. Since we’re multiplying z1, and z2, use De Moivre’s theorem by multiplying the r values (2 and 8) and adding the theta values (π/5 and 7π/6). Adding the angle values should lead you to have 41 π/30, and the rest is self-explanatory.

14. Explanation is in the answer, just take the r value from z2 for magnitude (at least that’s what’s on my practice assignment)

15. Use De Moivre’s theorem again, this time with division, so you will divide the r values (2 divided by 8) and subtract the theta values (π/5 minus 7π/6). 2/8 simplifies to 1/8 and when subtracting with 6π/30 - 35π/30 (finding common denominators) you should get 29π/30.

Answer 2

Explanation:

Given the following complex values Z₁=2(cos(π/5)+i Sin(πi/5)) And Z₂=8(cos(7π/6)+i Sin(7π/6)). We are to calculate the following complex numbers;

a) Z₁Z₂ = 2(cos(π/5)+i Sin(πi/5)) * 8(cos(7π/6)+i Sin(7π/6))

Z₁Z₂ = 18 {(cos(π/5)+i Sin(π/5))*(cos(7π/6)+i Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)+i²Sin(π/5)Sin(7π/6)) }

since i² = -1

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)-Sin(π/5)Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) -Sin(π/5)Sin(7π/6) + i(cos(π/5)sin(7π/6)+ Sin(π/5)cos(7π/6)) }

From trigonometry identity, cos(A+B) = cosAcosB - sinAsinB and sin(A+B) = sinAcosB + cosAsinB

The equation becomes

= 18{cos(π/5+7π/6) + isin(π/5+7π/6)) }

= 18{cos((6π+35π)/30) + isin(6π+35π)/30)) }

= 18{cos((41π)/30) + isin(41π)/30)) }

b) z2 value has already been given in polar form and it is equivalent to 8(cos(7pi/6)+i Sin(7pi/6))

c) for z1/z2 = 2(cos(pi/5)+i Sin(pi/5))/8(cos(7pi/6)+i Sin(7pi/6))

let A = pi/5 and B = 7pi/6

z1/z2 = 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B))

On rationalizing we will have;

= 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B)) * 8(cos(B)-i Sin(B))/8(cos(B)-i Sin(B))

= 16{cosAcosB-icosAsinB+isinAcosB-sinAsinB}/64{cos²B+sin²B}

= 16{cosAcosB-sinAsinB-i(cosAsinB-sinAcosB)}/64{cos²B+sin²B}

From trigonometry identity; cos²B+sin²B = 1

= 16{cos(A+ B)-i(sin(A+B)}/64

= 16{cos(pi/5+ 7pi/6)-i(sin(pi/5+7pi/6)}/64

= 16{ (cos 41π/30)-isin(41π/30)}/64

Z1/Z2 = (cos 41π/30)-isin(41π/30)/4