Question

Can someone please help me with this.

Answer 1

Explanation:

(1) see the picture

(2)

`\boxed{z=r(cos\theta+isin\theta)}`

where
`r=√(Re^2+Im^2)` and
`tan\theta=(Im)/(Re)`

`r=√(8^2+6^2)`

`=10`

`tan\theta=(6)/(8)`

`\theta=37^o`

Therefore, the trigonometric polar form for (6i + 8) = 10(cos 37° + i sin 37°)

(3)

z₁ · z₂ = r₁(cosθ₁ + isinθ₁) × r₂(cosθ₂ + isinθ₂)

= r₁ · r₂ · (cosθ₁ + isinθ₁)(cosθ₂ + isinθ₂)

= r₁ · r₂ · (cosθ₁cosθ₂ + isinθ₁cosθ₂ + isinθ₂cosθ₁ + isinθ₁isinθ₂)

= r₁ · r₂ · (cosθ₁cosθ₂ + i(sinθ₁cosθ₂ + sinθ₂cosθ₁) + i²sinθ₁sinθ₂)

= r₁ · r₂ · (cosθ₁cosθ₂ - sinθ₁sinθ₂+ i(sinθ₁cosθ₂ + cosθ₁sinθ₂))

= r₁ · r₂ · (cos(θ₁ + θ₂) + i(sin(θ₁ + θ₂)))

= r₁ · r₂ · cis(θ₁ + θ₂)

(4)

`(z_1)/(z_2) =(r_1(cos\theta_1+isin\theta_1))/(r_2(cos\theta_2+isin\theta_2))`

`=(r_1)/(r_2) \cdot((cos\theta_1+isin\theta_1))/((cos\theta_2+isin\theta_2)) \cdot((cos\theta_2-isin\theta_2))/((cos\theta_2-isin\theta_2))`

`=(r_1)/(r_2) \cdot(cos\theta_1+isin\theta_1)(cos\theta_2-isin\theta_2)`

`=(r_1)/(r_2) \cdot(cos\theta_1cos\theta_2+isin\theta_1cos\theta_2-isin\theta_2cos\theta_1-isin\theta_1isin\theta_2)`

`=(r_1)/(r_2) \cdot(cos\theta_1cos\theta_2-i^2sin\theta_1sin\theta_2+i(sin\theta_1cos\theta_2-sin\theta_2cos\theta_1))`

`=(r_1)/(r_2) \cdot((cos\theta_1cos\theta_2+sin\theta_1sin\theta_2)+i(sin\theta_1cos\theta_2-cos\theta_1sin\theta_2))`

`=(r_1)/(r_2) \cdot(cos(\theta_1-\theta_2)+i(sin(\theta_1-\theta_2)))`

`=(r_1)/(r_2) \cdot cis(\theta_1-\theta_2)`