Question

Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential form; (c) calculate A3 5 (A1 1A2), giving your answer in polar form; (d) calculate A4 5 A1A2, giving your answer in rectangular form; (e) calculate A5 5 A1ysA* 2d, giving your answer in exponential form.

Answer 1

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/(
`A^(*)`₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃ in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403
`e^(j51.34 )`

c) value of A₃ in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937
`e^(j81.34 )`

Step-by-step explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁ = 5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹(
`(5)/(4)` )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A
`e^(j\beta )`

so, A₂ in exponential form will be;

A₂ = 6.403
`e^(j51.34 )`

exponential form of A₂ = 6.403
`e^(j51.34 )`

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹(
`(8 )/(9.196)` )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°

A₃ = 12.19∠41.02°

Therefore, value of A₃ in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/(
`A^(*)`₂)

giving your answer in exponential form

we know that
`A^(*)`₂ is the complex conjugate of A₂

so

`A^(*)`₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A
`e^(j\beta )`

`A^(*)`₂ = 6.403
`e^(-j51.34 )`

also

A₁ = 6∠30

we convert to polar form

A₁ = 6
`e^(j30 )`

so A₅ = A₁/(
`A^(*)`₂)

A₅ = 6
`e^(j30 )` / 6.403
`e^(-j51.34 )`

A₅ = (6/6.403)
`e^(j(30+51.34) )`

A₅ = 0.937
`e^(j81.34 )`

Therefore A₅ in exponential form is 0.937
`e^(j81.34 )`