Let for real numbers and , and .
Find the number of values of satisfying for natural number .
Clarification: .
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We have ( z + z 2 ) ∣ z ∣ n − 2 = ∣ z ∣ n − 1 and hence z + z 2 is real. Thus the imaginary part of z + z 2 is equal to 0 , and hence y + 2 x y = ( 1 + 2 x ) y = 0 We are told that x = − 2 1 , so it follows that y = 0 , and hence z is real. Thus ∣ z ∣ n + z ∣ z ∣ n − 2 = ( z 2 + z ) ∣ z ∣ n − 2 = ∣ z ∣ n − 1 and hence z ∣ z ∣ n − 2 = − 1 , which implies that z = − 1 . There is only one possible value of z .