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/([tex]A^{*}[/tex]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[tex]e^{j51.34 }[/tex]
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[tex]e^{j81.34 }[/tex]
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( [tex]\frac{5}{4}[/tex] )
= √(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[tex]e^{j\beta }[/tex]
so, A₂ in exponential form will be;
A₂ = 6.403[tex]e^{j51.34 }[/tex]
exponential form of A₂ = 6.403[tex]e^{j51.34 }[/tex]
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( [tex]\frac{8 }{9.196}[/tex] )
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/([tex]A^{*}[/tex]₂)
giving your answer in exponential form
we know that [tex]A^{*}[/tex]₂ is the complex conjugate of A₂
so
[tex]A^{*}[/tex]₂ = (6.403 ∠51.34° )*
= 6.403 ∠-51.34°
we convert to exponential form
A∠β can be written as A[tex]e^{j\beta }[/tex]
[tex]A^{*}[/tex]₂ = 6.403[tex]e^{-j51.34 }[/tex]
also
A₁ = 6∠30
we convert to polar form
A₁ = 6[tex]e^{j30 }[/tex]
so A₅ = A₁/([tex]A^{*}[/tex]₂)
A₅ = 6[tex]e^{j30 }[/tex] / 6.403[tex]e^{-j51.34 }[/tex]
A₅ = (6/6.403) [tex]e^{j(30+51.34) }[/tex]
A₅ = 0.937[tex]e^{j81.34 }[/tex]
Therefore A₅ in exponential form is 0.937[tex]e^{j81.34 }[/tex]