Complex Condensation

Algebra Level 3

Given the following:

A ⃗ = A e j θ A B ⃗ = B e j θ B C ⃗ = C e j θ C D ⃗ = D e j θ D E ⃗ = E e j θ E ∣ A ⃗ ∣ + ∣ B ⃗ ∣ + ∣ C ⃗ ∣ = ∣ D ⃗ ∣ + ∣ E ⃗ ∣ A ⃗ + B ⃗ + C ⃗ = D ⃗ + E ⃗ \vec{A} = A \, e^{j \theta_A} \\ \vec{B} = B \, e^{j \theta_B} \\ \vec{C} = C \, e^{j \theta_C} \\ \vec{D} = D \, e^{j \theta_D} \\ \vec{E} = E \, e^{j \theta_E} \\ |\vec{A}| + |\vec{B}| + |\vec{C}| = |\vec{D}| + |\vec{E}| \\ \vec{A} + \vec{B} + \vec{C} = \vec{D} + \vec{E}

Determine the value of D D

Details and Assumptions:
1) j = − 1 j = \sqrt{-1}
2) ( A , B , C , D , E ) = ( 1 , 2 , 3 , ? , ? ) (A,B,C,D,E) = (1,2,3,?,?)
3) ( θ A , θ B , θ C , θ D , θ E ) = ( π 6 , π 4 , π 3 , 0 , ? ) (\theta_A,\theta_B,\theta_C, \theta_D, \theta_E) = (\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, 0, ?)
4) ∣ ⋅ ∣ |\cdot| denotes the magnitude of a complex number


The answer is 0.30387.

Steven Chase by Steven Chase , 37, USA

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1 solution

Nick Kent
Aug 10, 2019

Since ∣ e j ⋅ θ ∣ = 1 \left| { e }^{ j\cdot \theta } \right| =1 for any θ \theta , the magnitudes are easily found: ∣ A → ∣ = A = 1 , ∣ B → ∣ = B = 2 , ∣ C → ∣ = C = 3 , ∣ D → ∣ = D , ∣ E → ∣ = E \left| \overrightarrow { A } \right| =A=1,\left| \overrightarrow { B } \right| =B=2,\left| \overrightarrow { C } \right| =C=3,\left| \overrightarrow { D } \right| =D,\left| \overrightarrow { E } \right| =E . That means:

∣ A → ∣ + ∣ B → ∣ + ∣ C → ∣ = ∣ D → ∣ + ∣ E → ∣ 1 + 2 + 3 = D + E D + E = 6 \left| \overrightarrow { A } \right| +\left| \overrightarrow { B } \right| +\left| \overrightarrow { C } \right| =\left| \overrightarrow { D } \right| +\left| \overrightarrow { E } \right| \\ 1+2+3=D+E\\ D+E=6

Knowing that r e j θ = r cos ⁡ θ + j r sin ⁡ θ r{ e }^{ j\theta }=r\cos { \theta } +jr\sin { \theta } we can separate the real and imaginary parts of the last equation:

{ A cos ⁡ θ A + B cos ⁡ θ B + C cos ⁡ θ C = D cos ⁡ θ D + E cos ⁡ θ E A sin ⁡ θ A + B sin ⁡ θ B + C sin ⁡ θ C = D sin ⁡ θ D + E sin ⁡ θ E \begin{cases} A\cos { { \theta }_{ A } } +B\cos { { \theta }_{ B } } +C\cos { { \theta }_{ C } } =D\cos { { \theta }_{ D } } +E\cos { { \theta }_{ E } } \\ A\sin { { \theta }_{ A } } +B\sin { { \theta }_{ B } } +C\sin { { \theta }_{ C } } =D\sin { { \theta }_{ D } } +E\sin { { \theta }_{ E } } \end{cases}

Enter the values:

{ 1 cos ⁡ π 6 + 2 cos ⁡ π 4 + 3 cos ⁡ π 3 = D cos ⁡ 0 + E cos ⁡ θ E 1 sin ⁡ π 6 + 2 sin ⁡ π 4 + 3 sin ⁡ π 3 = D sin ⁡ 0 + E sin ⁡ θ E { 3 2 + 2 ⋅ 2 2 + 3 ⋅ 1 2 = D + E cos ⁡ θ E 1 2 + 2 ⋅ 2 2 + 3 ⋅ 3 2 = 0 + E sin ⁡ θ E { 3 2 + 2 + 3 2 = D + E cos ⁡ θ E 1 2 + 2 + 3 3 2 = E sin ⁡ θ E \begin{cases} 1\cos { \frac { \pi }{ 6 } } +2\cos { \frac { \pi }{ 4 } } +3\cos { \frac { \pi }{ 3 } } =D\cos { 0 } +E\cos { { \theta }_{ E } } \\ 1\sin { \frac { \pi }{ 6 } } +2\sin { \frac { \pi }{ 4 } } +3\sin { \frac { \pi }{ 3 } } =D\sin { 0 } +E\sin { { \theta }_{ E } } \end{cases}\\ \begin{cases} \frac { \sqrt { 3 } }{ 2 } +2\cdot \frac { \sqrt { 2 } }{ 2 } +3\cdot \frac { 1 }{ 2 } =D+E\cos { { \theta }_{ E } } \\ \frac { 1 }{ 2 } +2\cdot \frac { \sqrt { 2 } }{ 2 } +3\cdot \frac { \sqrt { 3 } }{ 2 } =0+E\sin { { \theta }_{ E } } \end{cases}\\ \begin{cases} \frac { \sqrt { 3 } }{ 2 } +\sqrt { 2 } +\frac { 3 }{ 2 } =D+E\cos { { \theta }_{ E } } \\ \frac { 1 }{ 2 } +\sqrt { 2 } +\frac { 3\sqrt { 3 } }{ 2 } =E\sin { { \theta }_{ E } } \end{cases}

Let u = 3 2 + 2 + 3 2 u=\frac { \sqrt { 3 } }{ 2 } +\sqrt { 2 } +\frac { 3 }{ 2 } and v = 1 2 + 2 + 3 3 2 v=\frac { 1 }{ 2 } +\sqrt { 2 } +\frac { 3\sqrt { 3 } }{ 2 } . Then:

{ D + E cos ⁡ θ E = u E sin ⁡ θ E = v { E cos ⁡ θ E = u − D E sin ⁡ θ E = v { E 2 ⋅ cos ⁡ 2 θ E = ( u − D ) 2 E 2 ⋅ sin ⁡ 2 θ E = v 2 \begin{cases} D+E\cos { { \theta }_{ E } } =u \\ E\sin { { \theta }_{ E } } =v \end{cases}\\ \begin{cases} E\cos { { \theta }_{ E } } =u-D \\ E\sin { { \theta }_{ E } } =v \end{cases}\\ \begin{cases} { E }^{ 2 }\cdot \cos ^{ 2 }{ { \theta }_{ E } } ={ \left( u-D \right) }^{ 2 } \\ { E }^{ 2 }\cdot \sin ^{ 2 }{ { \theta }_{ E } } ={ v }^{ 2 } \end{cases}

Let's sum the equations:

E 2 = ( u − D ) 2 + v 2 ( 6 − D ) 2 = ( u − D ) 2 + v 2 36 − 12 D + D 2 = u 2 − 2 u D + D 2 + v 2 ( 12 − 2 u ) D = 36 − u 2 − v 2 D = 36 − u 2 − v 2 12 − 2 u { E }^{ 2 }={ \left( u-D \right) }^{ 2 }+{ v }^{ 2 }\\ { \left( 6-D \right) }^{ 2 }={ \left( u-D \right) }^{ 2 }+{ v }^{ 2 }\\ 36-12D+{ D }^{ 2 }={ u }^{ 2 }-2uD+{ D }^{ 2 }+{ v }^{ 2 }\\ \left( 12-2u \right) D=36-{ u }^{ 2 }-{ v }^{ 2 }\\ D=\frac { 36-{ u }^{ 2 }-{ v }^{ 2 } }{ 12-2u }

Now we can evaluate D D :

D = 36 − u 2 − v 2 12 − 2 u = 36 − ( 3 2 + 2 + 3 2 ) 2 − ( 1 2 + 2 + 3 3 2 ) 2 12 − 2 ( 3 2 + 2 + 3 2 ) = 36 − 3 4 − 2 − 9 4 − 3 2 − 3 3 2 − 3 2 − 1 4 − 2 − 27 4 − 2 − 3 3 − 3 3 2 12 − 3 − 2 2 − 6 = 22 − 4 2 − 3 3 − 4 6 9 − 2 2 − 3 D=\frac { 36-{ u }^{ 2 }-{ v }^{ 2 } }{ 12-2u } =\frac { 36-{ \left( \frac { \sqrt { 3 } }{ 2 } +\sqrt { 2 } +\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } +\sqrt { 2 } +\frac { 3\sqrt { 3 } }{ 2 } \right) }^{ 2 } }{ 12-2\left( \frac { \sqrt { 3 } }{ 2 } +\sqrt { 2 } +\frac { 3 }{ 2 } \right) } =\frac { 36-\frac { 3 }{ 4 } -2-\frac { 9 }{ 4 } -\sqrt { 3 } \sqrt { 2 } -\frac { 3\sqrt { 3 } }{ 2 } -3\sqrt { 2 } -\frac { 1 }{ 4 } -2-\frac { 27 }{ 4 } -\sqrt { 2 } -3\sqrt { 3 } -3\sqrt { 3 } \sqrt { 2 } }{ 12-\sqrt { 3 } -2\sqrt { 2 } -6 } =\frac { 22-4\sqrt { 2 } -3\sqrt { 3 } -4\sqrt { 6 } }{ 9-2\sqrt { 2 } -\sqrt { 3 } }

So the answer is D = 0.30387 D=\boxed { 0.30387 }

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