Complex Square Root

Algebra Level 5

Find the value of k k that makes the equation

∣ ( p 4 + k p 2 q 2 + q 4 ) + ( 4 p 3 q − 4 p q 3 ) i ∣ = p 2 + q 2 \sqrt{\left|(p^4 + kp^2q^2 + q^4) + (4p^3q - 4pq^3)i\right|} = p^2 + q^2

true for all real values of p p and q q .


The answer is -6.

David Vreken by David Vreken , 42, USA

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2 solutions

David Vreken
Jan 5, 2020

Let a = p 4 + k p 2 q 2 + q 4 a = p^4 + kp^2q^2 + q^4 , b = 4 p 3 q − 4 p q 3 b = 4p^3q - 4pq^3 , and c = p 2 + q 2 c = p^2 + q^2 . Then the equation becomes ∣ a + b i ∣ = c \sqrt{|a + bi|} = c ,

Since ∣ a + b i ∣ = a 2 + b 2 |a + bi| = \sqrt{a^2 + b^2} , the equation becomes a 2 + b 2 = c \sqrt{\sqrt{a^2 + b^2}} = c , which means a 2 + b 2 = c 4 a^2 + b^2 = c^4 .

Substituting the original expressions back into a 2 + b 2 = c 4 a^2 + b^2 = c^4 gives:

( p 4 + k p 2 q 2 + q 4 ) 2 + ( 4 p 3 q − 4 p q 3 ) 2 = ( p 2 + q 2 ) 4 (p^4 + kp^2q^2 + q^4)^2 + (4p^3q - 4pq^3)^2 = (p^2 + q^2)^4

( p 8 + k 2 p 4 q 4 + q 8 + 2 k p 6 q 2 + 2 k p 2 q 6 + 2 p 4 q 4 ) + ( 16 p 6 q 2 − 32 p 4 q 4 + 16 p 2 q 6 ) = p 8 + 4 p 6 q 2 + 6 p 4 q 4 + 4 p 2 q 6 + q 8 (p^8 + k^2p^4q^4 + q^8 + 2kp^6q^2 + 2kp^2q^6 + 2p^4q^4) + (16p^6q^2 - 32p^4q^4 + 16p^2q^6) = p^8 + 4p^6q^2 + 6p^4q^4 + 4p^2q^6 + q^8

( 2 k + 12 ) p 6 q 2 + ( k 2 − 36 ) p 4 q 4 + ( 2 k + 12 ) p 2 q 6 = 0 (2k + 12)p^6q^2 + (k^2 - 36)p^4q^4 + (2k + 12)p^2q^6 = 0

2 ( k + 6 ) p 6 q 2 + ( k − 6 ) ( k + 6 ) p 4 q 4 + 2 ( k + 6 ) p 2 q 6 = 0 2(k + 6)p^6q^2 + (k - 6)(k + 6)p^4q^4 + 2(k + 6)p^2q^6 = 0

2 p 2 q 2 ( k + 6 ) ( p 4 + ( k − 6 ) 2 p 2 q 2 + q 4 ) = 0 2p^2q^2(k + 6)(p^4 + \frac{(k - 6)}{2}p^2q^2 + q^4) = 0

For this to be true for any value of p p and q q , k + 6 = 0 k + 6 = 0 which means k = − 6 k = \boxed{-6} .

2 Helpful 4 Interesting 0 Brilliant 1 Confused

Let p 4 + k p 2 q 2 + q 4 + 4 p q ( p 2 − q 2 ) = α + β i \sqrt{p^4+kp^2q^2 + q^4 + 4pq(p^2-q^2)} = \alpha + \beta i . Then we have:

{ α + β i = ( α + β i ) 2 = α 2 − β 2 + 2 α β i ⟹ { α 2 − β 2 = p 4 + k p 2 q 2 + q 4 . . . ( 1 ) α β = 2 p q ( p 2 − q 2 ) . . . ( 2 ) ∣ α + β i ∣ = p 2 + q 2 ⟹ α 2 + β 2 = p 2 + q 2 ⟹ α 2 + β 2 = p 4 + 2 p 2 q 2 + q 4 . . . ( 3 ) \begin{cases} \alpha + \beta i = \sqrt{(\alpha + \beta i)^2} = \sqrt{\alpha^2 - \beta^2 + 2\alpha \beta i} & \implies \begin{cases} \alpha^2 - \beta^2 = p^4+kp^2q^2 + q^4 & ... (1) \\ \alpha \beta = 2pq(p^2-q^2) & ...(2) \end{cases} \\ |\alpha + \beta i| = p^2 + q^2 \implies \sqrt{\alpha^2 + \beta^2} = p^2 + q^2 & \implies \alpha^2 + \beta^2 = p^4+2p^2q^2 + q^4 \qquad ... (3) \end{cases}

⟹ { ( 3 ) + ( 1 ) 2 : α 2 = p 4 + ( 2 + k 2 ) p 2 q 2 + q 4 . . . ( 4 ) ( 3 ) − ( 1 ) 2 : β 2 = ( 2 − k 2 ) p 2 q 2 . . . ( 5 ) \implies \begin{cases} \dfrac {(3)+(1)}2: & \alpha^2 = p^4+ \left(\dfrac {2+k}2\right) p^2q^2 + q^4 & ...(4) \\ \dfrac {(3)-(1)}2: & \beta^2 = \left(\dfrac {2-k}2\right) p^2q^2 & ...(5) \end{cases}

From ( 2 ) : α β = 2 p q ( p 2 − q 2 ) { α = p 4 + ( 2 + k 2 ) p 2 q 2 + q 4 = p 2 − q 2 β = 2 − k 2 p q = 2 p q ⟹ k = + 6 (2): \ \alpha \beta = 2pq(p^2-q^2) \begin{cases} \alpha = \sqrt{p^4+ \left(\dfrac {2+k}2\right) p^2q^2 + q^4} = p^2 - q^2 \\ \beta = \sqrt{\dfrac {2-k}2} pq = 2pq \end{cases} \implies k = \boxed{+6}

0 Helpful 0 Interesting 0 Brilliant 1 Confused

There are a few problems here. The biggest is that the square root function is not well-defined over the entire complex plane. Because, generally speaking, squaring is a 2-to-1 function and the easy algorithm of "take the positive root" that works over the reals cannot be extended analytically to the complex plane as a whole.

A second problem: if I let p=q=1, the imaginary terms disappear and we are left with 2 + k = 2. \sqrt{2+k} = 2. And that implies k = 2 k=2 .

But if I follow your logic, we end up with ∣ − 4 ∣ = 2 |\sqrt{-4}| = 2 . And that's the problem. Because − 4 \sqrt{-4} isn't really well-defined. There are two complex numbers that square to equal -4, and they both have norm 2. But, from the standpoint of rigor, this is a bit sketchy.

I looked at the left hand side as being essentially ( p + q i ) 4 \sqrt{ (p+qi)^4} , with k = − 6. k=-6. Indeed the system accepts k = − 6 k=-6 as an answer.

Did you make a typo when you said k = 6 k=6 ? Because while k = − 6 k=-6 is a problematic answer in some sense, k = 6 k=6 is surely false, and doesn't even work for your formula for α \alpha .

Richard Desper - 1 year, 5 months ago

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The answer should be k = − 6 k = -6 . There must be a typo in this solution.

De Moivre's Theorem is a well-defined method to take square roots on a complex plane and shows that there are always two answers with the same r r -value and therefore the same absolute value. For example, z = − 4 z = -4 converts to z = 4 ( cos ⁡ π + i sin ⁡ π ) z = 4(\cos \pi + i \sin \pi) , and its 2 2 second roots are 4 ( cos ⁡ π + 2 π 0 2 + i sin ⁡ π + 2 π 0 2 ) = 2 ( cos ⁡ π 2 + i sin ⁡ π 2 ) \sqrt{4}(\cos \frac{\pi + 2\pi0}{2} + i \sin \frac{\pi + 2\pi0}{2}) = 2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) and 4 ( cos ⁡ π + 2 π 1 2 + i sin ⁡ π + 2 π 1 2 ) = 2 ( cos ⁡ 3 π 2 + i sin ⁡ 3 π 2 ) \sqrt{4}(\cos \frac{\pi + 2\pi1}{2} + i \sin \frac{\pi + 2\pi1}{2}) = 2(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) , which converts back to 2 i 2i and − 2 i -2i , both of which have an absolute value of 0 2 + ( ± 2 ) 2 = 2 \sqrt{0^2 + (\pm 2)^2} = 2 .

Letting p = q = 1 p = q = 1 in the problem leads to ∣ 2 + k ∣ = 2 |\sqrt{2 + k}| = 2 (note the absolute value), which could mean k = 2 k = 2 or k = − 6 k = -6 . Some more examples of p p and q q would be needed to narrow down the answer, since the question says the equation is true for all real values of p p and q q .

David Vreken - 1 year, 5 months ago

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Again, the square root function is not well-defined on the complex plane. If somebody wants to define a "square root of the modulus" function, they are free to do so. That is something that is well-defined. But that's not what this problem is doing.

As you point out, there are always two solutions to z 2 = k z^2 = k for any complex number k ≠ 0 k \neq 0 . Your argument shows that − 4 \sqrt{-4} is not well defined. Actual functions don't have two values. That fact is the most important thing about functions.

"Letting p = q = 1 p = q = 1 in the problem leads to ∣ 2 + k ∣ = 2 |\sqrt{2 + k}| = 2 , which means k =4, since the absolute value function used outside the square root function is redundant. The square root function is only defined on non-negative real numbers and only returns non-negative values. ​
You and Chew-Seong have not been empowered to redefine the square-root function away from how it is understood to function in the world's mathematical community. Being "one of two roots of a second-degree equation" is a different thing than being "the value of a square root function" because, again, square root functions are single-valued! Moreover, it is impossible to define an analytic square root function the entire complex plane. One needs to create a branch cut to create a simply connected region that excludes zero. (Deleting the negative real axis would work, but that's not the only possibility.)

Richard Desper - 1 year, 5 months ago

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@Richard Desper – How about if I swap the square root and the absolute value symbols? Would that make it a better problem?

Like this:

∣ ( p 4 + k p 2 q 2 + q 4 ) + ( 4 p 3 q − 4 p q 3 ) i ∣ = p 2 + q 2 \sqrt{\left|(p^4 + kp^2q^2 + q^4) + (4p^3q - 4pq^3)i\right|} = p^2 + q^2

David Vreken - 1 year, 5 months ago

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@David Vreken – Yes, that would do it.

Richard Desper - 1 year, 5 months ago

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