Question

Consider equation and the closed circular region equation. Find points in R where |f(z)| has its maximum and minimum values.

09 Jan 2026
Answer :
Word Count : 477
We are asked to find the maximum and minimum of ( |f(z)| ) for ( f(z) = z^2 - z ) over the closed unit disk ( R = { z : |z| \le 1 } ). We solve this step by step manually. --- Let ( z = x + iy ), where ( x, y \in \mathbb{R} ). Then [ f(z) = z^2 - z = (x + iy)^2 - (x + iy) = (x^2 - y^2 - x) + i(2xy - y) ] So [ |f(z)|^2 = (x^2 - y^2 - x)^2 + (2xy - y)^2 ] Step 1: Check inside the disk (critical points) Critical points occur where the derivative ( f'(z) = 2z - 1 = 0 ), giving [ z = \frac{1}{2} ] This is inside ( R ) because ( |1/2|
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