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Timothy
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It's super easy to prove. For any complex number on the unit circle, you can findthere exists a sequence of complex numbers with integer parts whose direction approachessuch that if you replace each term with itself divided by its magnitude, the point half way alongnew sequence is a sequence of points on the unit circle from 1 to that pointapproaches the square root of the chosen complex number. ReplaceIf you replace each number in thatterm of the original sequence with its square and all numbers, then every term in thatthe sequence will have an integral magnitude so if. If after that, you then replace each number in thatterm of the new sequence with that numberitself divided by its magnitude, you getthe new sequence is a sequence of pointscomplex numbers on the unit circle with rational parts that approaches the chosen point on the unit circlecomplex number.

It's super easy to prove. For any complex number on the unit circle, you can find a sequence of complex numbers with integer parts whose direction approaches the point half way along the circle from 1 to that point. Replace each number in that sequence with its square and all numbers in that sequence have an integral magnitude so if you then replace each number in that sequence with that number divided by its magnitude, you get a sequence of points on the unit circle with rational parts that approaches the chosen point on the unit circle.

It's super easy to prove. For any complex number on the unit circle, there exists a sequence of complex numbers with integer parts such that if you replace each term with itself divided by its magnitude, the new sequence is a sequence of points on the unit circle that approaches the square root of the chosen complex number. If you replace each term of the original sequence with its square, then every term in the sequence will have an integral magnitude. If after that, you replace each term of the new sequence with itself divided by its magnitude, the new sequence is a sequence of complex numbers on the unit circle with rational parts that approaches the chosen complex number.

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Timothy
  • 803
  • 10
  • 17

It's super easy to prove. For any complex number on the unit circle, you can find a sequence of complex numbers with integer parts whose direction approaches the point half way along the circle from 1 to that point. Replace each number in that sequence with its square and all numbers in that sequence have an integral magnitude so if you then replace each number in that sequence with that number divided by its magnitude, you get a sequence of points on the unit circle with rational parts that approaches the chosen point on the unit circle.