Find the product of z₁ and z₂, where z₁ = 7(cos40° + i sin40°) and z₂ = 6(cos145° + i sin145°).

Solution:

Given: z\(_1\) = 7(cos40° + i sin40°) and z\(_2\)= 6(cos145° + i sin145°)

To find the product of the z\(_1\) and z\(_2\), we will use the rules of complex numbers.

For multiplying the complex numbers, that are in the polar form, when

z\(_1\) = r\(_1\) (cosθ + i sinθ)

z\(_2\) = r\(_2\)(cosϕ + i sinϕ )

we have, z\(_1\)× z\(_2\) = r\(_1\) × r\(_2\) (cos(θ + ϕ) + i sin(θ + ϕ)) --- equation(1)

Using equation(1),

r\(_1\) = 7, r\(_2\) = 6

θ = 40°, ϕ = 145°

z\(_1\)× z\(_2\) = 7 × 6(cos(40° + 145°) + i sin(40° + 145°))

z\(_1\)× z\(_2\) = 42(cos(185°) + i sin(185°))

Find the product of z₁ and z₂, where z₁ = 7(cos40° + i sin40°) and z₂ = 6(cos145° + i sin145°).

Summary:

The product of z₁ and z₂, where z₁ = 7(cos40° + i sin40°) and z₂ = 6(cos145° + i sin145°) is 42(cos(185°) + i sin(185°)).