Answer on Question #61150 - Math - Algebra
Question
Solve the equation
z 4 = − 2 3 − 2 i z ^ {4} = - 2 \sqrt {3} - 2 i z 4 = − 2 3 − 2 i
Sketch the solution in the complex plane.
Solution
The absolute value (the modulus) of the complex number − 2 3 − 2 i -2\sqrt{3} - 2i − 2 3 − 2 i
ρ ( z 4 ) = ( − 2 3 ) 2 + ( − 2 ) 2 = 4 \rho (z ^ {4}) = \sqrt {\left(- 2 \sqrt {3}\right) ^ {2} + (- 2) ^ {2}} = 4 ρ ( z 4 ) = ( − 2 3 ) 2 + ( − 2 ) 2 = 4
The argument of the complex number − 2 3 − 2 i -2\sqrt{3} - 2i − 2 3 − 2 i
θ ( z 4 ) = − π + tan − 1 ( − 2 − 2 3 ) = − π + π 6 = − 5 π 6 o r 2 π − 5 π 6 = 7 π 6 . \theta (z ^ {4}) = - \pi + \tan^ {- 1} \left(\frac {- 2}{- 2 \sqrt {3}}\right) = - \pi + \frac {\pi}{6} = - \frac {5 \pi}{6} \mathrm {o r} 2 \pi - \frac {5 \pi}{6} = \frac {7 \pi}{6}. θ ( z 4 ) = − π + tan − 1 ( − 2 3 − 2 ) = − π + 6 π = − 6 5 π or 2 π − 6 5 π = 6 7 π .
The absolute value (the modulus) of the complex number z = − 2 3 − 2 i 4 z = \sqrt[4]{-2\sqrt{3} - 2i} z = 4 − 2 3 − 2 i
ρ ( z ) = 4 4 = 2 ≈ 1.41 \rho (z) = \sqrt [ 4 ]{4} = \sqrt {2} \approx 1. 4 1 ρ ( z ) = 4 4 = 2 ≈ 1.41
The argument of the complex number z = − 2 3 − 2 i 4 z = \sqrt[4]{-2\sqrt{3} - 2i} z = 4 − 2 3 − 2 i
θ ( z ) = 7 π 6 + 2 π k 4 , k = 0 , 1 , 2 , 3. \theta (z) = \frac {\frac {7 \pi}{6} + 2 \pi k}{4}, k = 0, 1, 2, 3. θ ( z ) = 4 6 7 π + 2 πk , k = 0 , 1 , 2 , 3.
Solutions to the equation (1) are given by
z = ρ ( z ) ⋅ ( cos θ ( z ) + i sin θ ( z ) ) , z = \rho (z) \cdot (\cos \theta (z) + i \sin \theta (z)), z = ρ ( z ) ⋅ ( cos θ ( z ) + i sin θ ( z )) ,
where ρ ( z ) \rho (z) ρ ( z ) θ ( z ) \theta (z) θ ( z )
If k = 0 k = 0 k = 0 θ ( z ) = θ 1 = 7 π 6 4 = 7 π 24 ≈ 52.5 ∘ \theta(z) = \theta_1 = \frac{\frac{7\pi}{6}}{4} = \frac{7\pi}{24} \approx 52.5{}^\circ θ ( z ) = θ 1 = 4 6 7 π = 24 7 π ≈ 52.5 ∘
cos 7 π 24 = 1 2 2 − 2 − 3 ≈ 0.61 \cos \frac{7\pi}{24} = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{3}}} \approx 0.61 cos 24 7 π = 2 1 2 − 2 − 3 ≈ 0.61
sin 7 π 24 = 1 2 2 + 2 − 3 ≈ 0.79 \sin \frac{7\pi}{24} = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{3}}} \approx 0.79 sin 24 7 π = 2 1 2 + 2 − 3 ≈ 0.79
and a solution will be
z = z 1 = ρ ( z ) ⋅ ( cos θ 1 + i sin θ 1 ) = 2 ( cos 7 π 24 + i sin 7 π 24 ) ≈ 0.86 + 1.12 i . z = z _ {1} = \rho (z) \cdot (\cos \theta_ {1} + i \sin \theta_ {1}) = \sqrt {2} \left(\cos \frac {7 \pi}{2 4} + i \sin \frac {7 \pi}{2 4}\right) \approx 0. 8 6 + 1. 1 2 i. z = z 1 = ρ ( z ) ⋅ ( cos θ 1 + i sin θ 1 ) = 2 ( cos 24 7 π + i sin 24 7 π ) ≈ 0.86 + 1.12 i .
If k = 1 k = 1 k = 1 θ ( z ) = θ 2 = 7 π 6 + 2 π 4 = 19 π 24 ≈ 142.5 ∘ \theta(z) = \theta_2 = \frac{\frac{7\pi}{6} + 2\pi}{4} = \frac{19\pi}{24} \approx 142.5{}^\circ θ ( z ) = θ 2 = 4 6 7 π + 2 π = 24 19 π ≈ 142.5 ∘
cos 19 π 24 = cos ( π − 5 π 24 ) = − cos 5 π 24 = − 1 2 2 + 2 − 3 ≈ − 0.79 \cos \frac {1 9 \pi}{2 4} = \cos \left(\pi - \frac {5 \pi}{2 4}\right) = - \cos \frac {5 \pi}{2 4} = - \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx - 0. 7 9 cos 24 19 π = cos ( π − 24 5 π ) = − cos 24 5 π = − 2 1 2 + 2 − 3 ≈ − 0.79
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
sin 19 π 24 = sin ( π − 5 π 24 ) = sin 5 π 24 = 1 2 2 − 2 − 3 ≈ 0.61 \sin \frac {19\pi}{24} = \sin \left(\pi - \frac {5\pi}{24}\right) = \sin \frac {5\pi}{24} = \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx 0.61 sin 24 19 π = sin ( π − 24 5 π ) = sin 24 5 π = 2 1 2 − 2 − 3 ≈ 0.61
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
and a solution will be
z = z 2 = ρ ( z ) ⋅ ( cos θ 2 + i sin θ 2 ) = 2 ( cos 19 π 24 + i sin 19 π 24 ) ≈ − 1.12 + 0.86 i . z = z _ {2} = \rho (z) \cdot (\cos \theta_ {2} + i \sin \theta_ {2}) = \sqrt {2} \left(\cos \frac {19\pi}{24} + i \sin \frac {19\pi}{24}\right) \approx - 1.12 + 0.86i. z = z 2 = ρ ( z ) ⋅ ( cos θ 2 + i sin θ 2 ) = 2 ( cos 24 19 π + i sin 24 19 π ) ≈ − 1.12 + 0.86 i .
If k = 2 k = 2 k = 2 θ ( z ) = θ 3 = 7 π 6 + 4 π 4 = 31 π 24 ≈ 232.5 ∘ \theta (z) = \theta_{3} = \frac{\frac{7\pi}{6} + 4\pi}{4} = \frac{31\pi}{24}\approx 232.5{}^{\circ} θ ( z ) = θ 3 = 4 6 7 π + 4 π = 24 31 π ≈ 232.5 ∘
cos 31 π 24 = cos ( π + 7 π 24 ) = − cos 7 π 24 = − 1 2 2 − 2 − 3 ≈ − 0.61 \cos \frac {31\pi}{24} = \cos \left(\pi + \frac {7\pi}{24}\right) = - \cos \frac {7\pi}{24} = - \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx - 0.61 cos 24 31 π = cos ( π + 24 7 π ) = − cos 24 7 π = − 2 1 2 − 2 − 3 ≈ − 0.61
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
sin 31 π 24 = sin ( π + 7 π 24 ) = − sin 7 π 24 = − 1 2 2 + 2 − 3 ≈ − 0.79 \sin \frac {31\pi}{24} = \sin \left(\pi + \frac {7\pi}{24}\right) = - \sin \frac {7\pi}{24} = - \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx - 0.79 sin 24 31 π = sin ( π + 24 7 π ) = − sin 24 7 π = − 2 1 2 + 2 − 3 ≈ − 0.79
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
and a solution will be
z = z 3 = ρ ( z ) ⋅ ( cos θ 3 + i sin θ 3 ) = 2 ( cos 31 π 24 + i sin 31 π 24 ) ≈ − 0.86 − 1.12 i z = z _ {3} = \rho (z) \cdot (\cos \theta_ {3} + i \sin \theta_ {3}) = \sqrt {2} \left(\cos \frac {31\pi}{24} + i \sin \frac {31\pi}{24}\right) \approx - 0.86 - 1.12i z = z 3 = ρ ( z ) ⋅ ( cos θ 3 + i sin θ 3 ) = 2 ( cos 24 31 π + i sin 24 31 π ) ≈ − 0.86 − 1.12 i
If k = 3 k = 3 k = 3 θ ( z ) = θ 4 = 7 π 6 + 6 π 4 = 43 π 24 ≈ 322.5 ∘ \theta (z) = \theta_{4} = \frac{\frac{7\pi}{6} + 6\pi}{4} = \frac{43\pi}{24}\approx 322.5{}^{\circ} θ ( z ) = θ 4 = 4 6 7 π + 6 π = 24 43 π ≈ 322.5 ∘
cos 43 π 24 = cos ( 48 π − 5 π 24 ) = cos ( 2 π − 5 π 24 ) = cos ( − 5 π 24 ) = cos 5 π 24 = 1 2 2 + 2 − 3 ≈ 0.79 \cos \frac {43\pi}{24} = \cos \left(\frac {48\pi - 5\pi}{24}\right) = \cos \left(2\pi - \frac {5\pi}{24}\right) = \cos \left(- \frac {5\pi}{24}\right) = \cos \frac {5\pi}{24} = \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx 0.79 cos 24 43 π = cos ( 24 48 π − 5 π ) = cos ( 2 π − 24 5 π ) = cos ( − 24 5 π ) = cos 24 5 π = 2 1 2 + 2 − 3 ≈ 0.79
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
sin 43 π 24 = sin ( 48 π − 5 π 24 ) = sin ( 2 π − 5 π 24 ) = sin ( − 5 π 24 ) = − sin 5 π 24 = − 1 2 2 − 2 − 3 ≈ − 0.61 \sin \frac {43\pi}{24} = \sin \left(\frac {48\pi - 5\pi}{24}\right) = \sin \left(2\pi - \frac {5\pi}{24}\right) = \sin \left(- \frac {5\pi}{24}\right) = - \sin \frac {5\pi}{24} = - \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx - 0.61 sin 24 43 π = sin ( 24 48 π − 5 π ) = sin ( 2 π − 24 5 π ) = sin ( − 24 5 π ) = − sin 24 5 π = − 2 1 2 − 2 − 3 ≈ − 0.61
(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),
and a solution will be
z = z 4 = ρ ( z ) ⋅ ( cos θ 4 + i sin θ 4 ) = 2 ( cos 43 π 24 + i sin 43 π 24 ) ≈ 1.12 − 0.86 i z = z _ {4} = \rho (z) \cdot (\cos \theta_ {4} + i \sin \theta_ {4}) = \sqrt {2} \left(\cos \frac {43\pi}{24} + i \sin \frac {43\pi}{24}\right) \approx 1.12 - 0.86i z = z 4 = ρ ( z ) ⋅ ( cos θ 4 + i sin θ 4 ) = 2 ( cos 24 43 π + i sin 24 43 π ) ≈ 1.12 − 0.86 i
Table 1. Arguments of the solutions
Figure 1. The solutions to the equation z 4 = − 2 3 − 2 i z^4 = -2\sqrt{3} - 2i z 4 = − 2 3 − 2 i
Now show how to deduce auxiliary formulae, for example,
cos 5 π 24 = 1 2 2 + 2 − 3 , sin 5 π 24 = 1 2 2 − 2 − 3 \cos \frac{5\pi}{24} = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{3}}}, \sin \frac{5\pi}{24} = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{3}}} cos 24 5 π = 2 1 2 + 2 − 3 , sin 24 5 π = 2 1 2 − 2 − 3
It is known that cos 30 ∘ = cos π 6 = 3 2 \cos 30{}^{\circ} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} cos 30 ∘ = cos 6 π = 2 3 sin 30 ∘ = sin π 6 = 1 2 \sin 30{}^{\circ} = \sin \frac{\pi}{6} = \frac{1}{2} sin 30 ∘ = sin 6 π = 2 1
Using half-angle formulae
sin 2 π 24 = sin 15 ∘ = sin 30 ∘ 2 = 1 − cos 30 ∘ 2 = 1 − 3 2 2 = 1 2 2 − 3 , \sin \frac{2\pi}{24} = \sin 15{}^{\circ} = \sin \frac{30{}^{\circ}}{2} = \sqrt{\frac{1 - \cos 30{}^{\circ}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2}\sqrt{2 - \sqrt{3}}, sin 24 2 π = sin 15 ∘ = sin 2 30 ∘ = 2 1 − c o s 30 ∘ = 2 1 − 2 3 = 2 1 2 − 3 ,
cos 2 π 24 = cos 15 ∘ = cos 30 ∘ 2 = 1 + cos 30 ∘ 2 = 1 + 3 2 2 = 1 2 2 + 3 . \cos \frac{2\pi}{24} = \cos 15{}^{\circ} = \cos \frac{30{}^{\circ}}{2} = \sqrt{\frac{1 + \cos 30{}^{\circ}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2}\sqrt{2 + \sqrt{3}}. cos 24 2 π = cos 15 ∘ = cos 2 30 ∘ = 2 1 + c o s 30 ∘ = 2 1 + 2 3 = 2 1 2 + 3 .
Using reduction formulae
cos 75 ∘ = sin ( 90 ∘ − 75 ∘ ) = sin 15 ∘ = 1 2 2 − 3 \cos 75{}^{\circ} = \sin \left(90{}^{\circ} - 75{}^{\circ}\right) = \sin 15{}^{\circ} = \frac{1}{2}\sqrt{2 - \sqrt{3}} cos 75 ∘ = sin ( 90 ∘ − 75 ∘ ) = sin 15 ∘ = 2 1 2 − 3
sin 75 ∘ = sin ( 90 ∘ − 15 ∘ ) = cos 15 ∘ = 1 2 2 + 3 \sin 75{}^{\circ} = \sin \left(90{}^{\circ} - 15{}^{\circ}\right) = \cos 15{}^{\circ} = \frac{1}{2}\sqrt{2 + \sqrt{3}} sin 75 ∘ = sin ( 90 ∘ − 15 ∘ ) = cos 15 ∘ = 2 1 2 + 3
Using half-angle formulae finally obtain
sin 37.5 ∘ = sin 5 π 24 = sin ( 5 π / 12 2 ) = sin ( 75 ∘ 2 ) = 1 − cos 75 ∘ 2 = 1 − 1 2 2 − 3 2 = 1 2 2 − 2 − 3 , \sin 37.5{}^{\circ} = \sin \frac{5\pi}{24} = \sin \left(\frac{5\pi/12}{2}\right) = \sin \left(\frac{75{}^{\circ}}{2}\right) = \sqrt{\frac{1 - \cos 75{}^{\circ}}{2}} = \sqrt{\frac{1 - \frac{1}{2}\sqrt{2 - \sqrt{3}}}{2}} = \frac{1}{2} \sqrt{2 - \sqrt{2 - \sqrt{3}}}, sin 37.5 ∘ = sin 24 5 π = sin ( 2 5 π /12 ) = sin ( 2 75 ∘ ) = 2 1 − cos 75 ∘ = 2 1 − 2 1 2 − 3 = 2 1 2 − 2 − 3 , cos 37.5 ∘ = cos 5 π 24 = cos ( 5 π / 12 2 ) = cos ( 75 ∘ 2 ) = 1 + cos 75 ∘ 2 = 1 + 1 2 2 − 3 2 = 1 2 2 + 2 − 3 . \cos 37.5{}^{\circ} = \cos \frac{5\pi}{24} = \cos \left(\frac{5\pi/12}{2}\right) = \cos \left(\frac{75{}^{\circ}}{2}\right) = \sqrt{\frac{1 + \cos 75{}^{\circ}}{2}} = \sqrt{\frac{1 + \frac{1}{2}\sqrt{2 - \sqrt{3}}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{2 - \sqrt{3}}}. cos 37.5 ∘ = cos 24 5 π = cos ( 2 5 π /12 ) = cos ( 2 75 ∘ ) = 2 1 + cos 75 ∘ = 2 1 + 2 1 2 − 3 = 2 1 2 + 2 − 3 .
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#340153 on Dec 2023