An algebra problem by Keshav Tiwari

Algebra Level 5

Let z = x + i y z = x + iy for real numbers x x and y y , and x 1 2 x \ne -\dfrac12 .

Find the number of values of z z satisfying z n = z 2 z n 2 + z z n 2 + 1 |z|^{n} = z^{2} |z|^{n-2} +z|z|^{n-2} +1 for natural number n > 1 n> 1 .

Clarification: i = 1 i=\sqrt{-1} .

1 3 2 0

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1 solution

Mark Hennings
Nov 8, 2016

We have ( z + z 2 ) z n 2 = z n 1 (z + z^2)|z|^{n-2} \; = \; |z|^n - 1 and hence z + z 2 z + z^2 is real. Thus the imaginary part of z + z 2 z + z^2 is equal to 0 0 , and hence y + 2 x y = ( 1 + 2 x ) y = 0 y + 2xy \; = \; (1 + 2x)y \; = \; 0 We are told that x 1 2 x \neq -\tfrac12 , so it follows that y = 0 y=0 , and hence z z is real. Thus z n + z z n 2 = ( z 2 + z ) z n 2 = z n 1 |z|^n + z|z|^{n-2} \; = \; (z^2+z)|z|^{n-2} \; = \; |z|^n - 1 and hence z z n 2 = 1 z|z|^{n-2} = -1 , which implies that z = 1 z=-1 . There is only one possible value of z z .

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