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Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex numbers.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

Function of Complex Numbers – AIME I, 1999


Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that \(b^{2}\)=\(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 259
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Integers

Complex Numbers

Check the Answer


Answer: is 259.

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through \((\frac{1}{2},\frac{1}{2})\) that is x+y=1

Second Hint

putting x=(a-b) and y=(a+b) gives 2a=1 and \(a=\frac{1}{2}\)

Final Step

and \((\frac{1}{2})^{2} +b^{2}=8^{2}\) then \(b^{2}=\frac{255}{4}\) then 255+4=259.

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