Given that there is a complex number z such that ∣ ∣ ∣ ∣ z + z 1 ∣ ∣ ∣ ∣ = 2 , find the maximum possible value of ∣ z ∣ .
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We have ( z + z − 1 ) ( z + z − 1 ) ∣ z ∣ 2 + ∣ z ∣ − 2 + z z − 1 + z z − 1 = 4 = 4 Let w = z z − 1 . Then ∣ w ∣ = 1 and the equation becomes ∣ z ∣ 2 + ∣ z ∣ − 2 + 2 Re ( w ) ∣ z ∣ 2 + ∣ z ∣ − 2 = 4 = 4 − 2 Re ( w ) . We want to maximize the left side of this equation, as it is an increasing function of ∣ z ∣ for ∣ z ∣ > 1 . Since Re ( w ) is between − 1 and 1 , the right side is at most 6 . So the maximum occurs when it is 6 , which is when w = − 1 , or z = − z , or z = r i for some real number r . In that case, we get ∣ z ∣ 2 + ∣ z ∣ − 2 ∣ z ∣ 4 − 6 ∣ z ∣ 2 + 1 = 6 = 0 and solving via the quadratic equation gives ∣ z ∣ 2 = 3 ± 2 2 . The plus sign gives a larger ∣ z ∣ , and the positive square root of 3 + 2 2 is 1 + 2 , so the answer is 2 . 4 1 4 .
Put z = x + i y and rewrite the given expression as ∣ ∣ z 2 + 1 ∣ ∣ = 2 ∣ z ∣ . Squaring both sides and rearranging, ( x 2 − y 2 + 1 ) 2 + 4 x 2 y 2 ( x 2 − y 2 ) 2 + 2 ( x 2 − y 2 ) + 1 + 4 x 2 y 2 ( x 2 + y 2 ) 2 + 2 ( x 2 − y 2 ) + 1 ( x 2 + y 2 ) 2 − 2 ( x 2 + y 2 ) + 1 ( x 2 + y 2 − 1 ) 2 ( x 2 + y 2 − 1 − 2 y ) ( x 2 + y 2 − 1 + 2 y ) = 4 ( x 2 + y 2 ) = 4 ( x 2 + y 2 ) = 4 ( x 2 + y 2 ) = 4 y 2 = ( 2 y ) 2 = 0
where the last line comes from rewriting as a difference of two squares. Completing the square in each factor,
( x 2 + ( y − 1 ) 2 − 2 ) ( x 2 + ( y + 1 ) 2 − 2 ) = 0
For the product to be zero, (at least) one of the factors has to be zero. Equating the red factor to zero gives a circle centred at
(
0
,
1
)
with radius
2
; equating the blue factor to zero gives a circle centred at
(
0
,
−
1
)
with radius
2
. So the locus of points satisfying the original condition is just two intersecting circles, as below:
It's clear from the geometry that ∣ z ∣ is maximised when z = ( 1 + 2 ) i , ie ∣ z ∣ = 1 + 2 .
One final point: it seems there may be a shortcut to this solution, without rewriting z in terms of reals, if we note that ∣ z 2 + 1 ∣ = ∣ z − i ∣ ⋅ ∣ z + i ∣ . I haven't managed to find one, though - any ideas?
I just posted a (mostly) "coordinate-free" solution.
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Nice (although saying Re ( w ) isn't a coordinate stretches a point ;-)). I wondered, though, is there any way of showing that the locus is two linked circles without coordinates? (To me, that's the nice thing about this problem - I was very surprised that locus came up from such an innocuous equation)
Finally had a chance to look at this. Here you go: ∣ ∣ ∣ ∣ z + z 1 ∣ ∣ ∣ ∣ ∣ ∣ z 2 + 1 ∣ ∣ ∣ z ∣ 4 + z 2 + z 2 + 1 ∣ z ∣ 4 − 2 ∣ z ∣ 2 + 1 + z 2 + z 2 − 2 ∣ z ∣ 2 ( ∣ z ∣ 2 − 1 ) 2 + ( z − z ) 2 ( ∣ z ∣ 2 − 1 + i ( z − z ) ) ( ∣ z ∣ 2 − 1 − i ( z − z ) ) ( ( z − i ) ( z + i ) − 2 ) ( ( z + i ) ( z − i ) − 2 ) ( ∣ z − i ∣ 2 − 2 ) ( ∣ z + i ∣ 2 − 2 ) ∣ z − i ∣ = 2 = 2 = 2 ∣ z ∣ = 4 ∣ z ∣ 2 = 0 = 0 = 0 = 0 = 0 or ∣ z + i ∣ = 2 and the steps are reversible.
Let z = x + i y . Then ∣ z + z 1 ∣ = 2 ⟹ x 4 + 2 x 2 y 2 + y 4 − 2 x 2 − 6 y 2 + 1 = 0 ⟹ 4 x 3 d x + 4 x y 2 d x + 4 x 2 y d y + 4 y 3 d y − 4 x d x − 1 2 y d y = 0 . ∣ z ∣ = x 2 + y 2 = maximum ⟹ 2 x d x + 2 y d y = 0 ⟹ d y = − y x d x ⟹ 4 x 3 d x + 4 x y 2 d x − 4 x 3 d x − 4 x y 2 d x − 4 x d x + 1 2 x d x = 0 ⟹ x = 0 ⟹ y 4 − 6 y 2 + 1 = 0 ⟹ y 2 = 3 + 2 2 ⟹ ∣ z ∣ m a x = 3 + 2 2 = 2 + 1 ≈ 2 . 4 1 4 2 1 3 5 6 2 3 7 3 1 .
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Nice solutions, everyone! Here's another solution: ∣ z ∣ = ∣ ∣ ∣ ∣ z + z 1 − z 1 ∣ ∣ ∣ ∣ ≤ ∣ ∣ ∣ ∣ z + z 1 ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ z 1 ∣ ∣ ∣ ∣ = 2 + ∣ z ∣ 1 . The rest should be trivial. I liked this solution because there was no bash. ^_^