How complexity does a number achieve? #3

Algebra Level 3

If the 6 6 solutions of x 6 = − 64 x^{6}=-64 are written in the form a + i b a+ib , where a a and b b are real, then what is the product of those solutions with a > 0 a>0 ?


The answer is 4.

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Parag Zode by Parag Zode , 24, India

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3 solutions

We have that

x 6 + 64 = ( x 2 + 4 ) ( x 4 − 4 x 2 + 16 ) = x^{6} + 64 = (x^{2} + 4)(x^{4} - 4x^{2} + 16) =

( x − 2 i ) ( x + 2 i ) ( x 2 − ( 2 + 2 3 i ) ) ( x 2 − ( 2 − 2 3 i ) ) = (x - 2i)(x + 2i)(x^{2} - (2 + 2\sqrt{3}i))(x^{2} - (2 - 2\sqrt{3}i)) =

( x − 2 i ) ( x + 2 i ) ( x − 2 + 2 3 i ) ( x + 2 + 2 3 i ) ( x − 2 − 2 3 i ) ( x + 2 − 2 3 i ) . (x - 2i)(x + 2i)(x - \sqrt{2 + 2\sqrt{3}i})(x + \sqrt{2 + 2\sqrt{3}i})(x - \sqrt{2 - 2\sqrt{3}i})(x + \sqrt{2 - 2\sqrt{3}i}).

The only roots that have a > 0 a \gt 0 are

2 + 2 3 i = 3 + i \sqrt{2 + 2\sqrt{3}i} = \sqrt{3} + i and 2 − 2 3 i = 3 − i \sqrt{2 - 2\sqrt{3}i} = \sqrt{3} - i ,

which have a product of 3 − i 2 = 4 . 3 - i^{2} = \boxed{4}.

20 Helpful 0 Interesting 0 Brilliant 0 Confused

We don't even need to calculate the roots individually like that. After factoring first, we can see that x = 2 i , ( − 2 i ) x=2i,(-2i) aren't required solutions since their real parts is 0 0 . Now, solving x 4 − 4 x 2 + 16 = 0 x^4-4x^2+16=0 using quadratic formula gives us,

x 2 = 4 ± 4 3 i 2 ⟹ x 2 4 = e π 3 i , e 5 π 3 i ⟹ x = ± 2 e π 6 i , ± 2 e 5 π 6 i \large x^2=\frac{4\pm 4\sqrt{3}i}{2}\implies \frac{x^2}{4}=e^{\frac{\pi}{3}i}~,~e^{\frac{5\pi}{3}i}\\ \implies \large x=\pm 2e^{\frac{\pi}{6}i}~,~\pm 2e^{\frac{5\pi}{6}i}

Examining the real parts of the roots, i.e., the cosine of the angles of the roots in polar form and using the fact that the cosine of only first and fourth quadrant angles are positive and for the remaining quadrant angles, it is negative, tells us that the only possible solutions that fit the problem criterions are,

x = 2 e π 6 i , − 2 e 5 π 6 i \large x=2e^{\frac{\pi}{6}i}~,~-2e^{\frac{5\pi}{6}i}

Let the required product be X X . We then have,

X = ( − 4 ) × e ( π 6 + 5 π 6 ) i = ( − 4 ) e i π = ( − 4 ) ( cos ⁡ π + i sin ⁡ π ) = 4 \large X=(-4)\times e^{\left(\frac{\pi}{6}+\frac{5\pi}{6}\right)i}=(-4)e^{i\pi}=(-4)(\cos\pi+i\sin\pi)=\boxed{4}

Prasun Biswas - 6 years, 2 months ago

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Haven't thought that it can be done by simply using quadratic formula.. Great observation @Prasu

Parag Zode - 6 years, 2 months ago

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Thanks. :)

Prasun Biswas - 6 years, 2 months ago
Chew-Seong Cheong
Mar 22, 2015

I used the method used by Mariano PerezdelaCruz .

x 6 = − 64 = 2 6 ( − 1 ) = 2 6 ( e ( 2 k + 1 ) π i ) x^6 = -64 = 2^6(-1)=2^6\left(e^{(2k+1)\pi i} \right) , where k = 0 , 1 , 2 , 3 , 4 , 5 k=0,1,2,3,4,5 .

⇒ x = 2 e 2 k + 1 6 π i = { 2 ( cos ⁡ π 6 + i sin ⁡ π 6 ) = 3 + i 2 ( cos ⁡ π 2 + i sin ⁡ π 2 ) = 0 + 2 i 2 ( cos ⁡ 5 π 6 + i sin ⁡ 5 π 6 ) = − 3 + i 2 ( cos ⁡ 7 π 6 + i sin ⁡ 7 π 6 ) = − 3 − i 2 ( cos ⁡ 3 π 2 + i sin ⁡ 3 π 2 ) = 0 − 2 i 2 ( cos ⁡ 11 π 6 + i sin ⁡ 11 π 6 ) = 3 − i \Rightarrow x = 2e^{\frac{2k+1}{6}\pi i} = \begin{cases} 2(\cos {\frac {\pi}{6}} + i\sin {\frac {\pi}{6}}) & = \sqrt{3}+i \\ 2(\cos {\frac {\pi}{2}} + i\sin {\frac {\pi}{2}}) & = 0+2i \\ 2(\cos {\frac {5\pi}{6}} + i\sin {\frac {5\pi}{6}}) & = -\sqrt{3}+i \\ 2(\cos {\frac {7\pi}{6}} + i\sin {\frac {7\pi}{6}}) & = -\sqrt{3}-i \\ 2(\cos {\frac {3\pi}{2}} + i\sin {\frac {3\pi}{2}}) & = 0-2i \\ 2(\cos {\frac {11\pi}{6}} + i\sin {\frac {11 \pi}{6}}) & = \sqrt{3}-i \end{cases}

The product of solutions with a > 0 a>0 is = ( 3 + i ) ( 3 − i ) = 4 = (\sqrt{3}+i)(\sqrt{3}-i) = \boxed{4} .

13 Helpful 0 Interesting 0 Brilliant 1 Confused
Geoff Taylor
Jan 9, 2018

x 6 = 2 6 ( c i s ( π + 2 k π ) ) , k ∈ Z { x }^{ 6 }={ 2 }^{ 6 }(cis(\pi +2k\pi )),\quad k\in Z

Thus x = 2 c i s ( 2 k + 1 6 π ) x=2cis(\frac { 2k+1 }{ 6 } \pi )

assigning 6 values for k ...0, 1, 2, 3, 4, 5

x = 2 c i s π 6 , 2 c i s 3 π 6 , 2 c i s 5 π 6 , 2 c i s 7 π 6 , 2 c i s 9 π 6 , 2 c i s 11 π 6 x=2cis\frac { \pi }{ 6 } ,2cis\frac { 3\pi }{ 6 } ,2cis\frac { 5\pi }{ 6 } ,2cis\frac { 7\pi }{ 6 } ,2cis\frac { 9\pi }{ 6 } ,2cis\frac { 11\pi }{ 6 }

the only values of x for which a >0 from these are x = 2 c i s π 6 , 2 c i s 11 π 6 x=2cis\frac { \pi }{ 6 } ,2cis\frac { 11\pi }{ 6 }

The product of these is 2 c i s π 6 × 2 c i s 11 π 6 = 4 c i s 2 π = ∗ ∗ 4 ∗ ∗ 2cis\frac { \pi }{ 6 } \times 2cis\frac { 11\pi }{ 6 } =4cis2\pi =**4**

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