My Math Forum |z² - 2z -1| = 0 (Complex Variable)?

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 February 3rd, 2017, 11:45 AM #1 Newbie   Joined: Jan 2017 From: United States Posts: 4 Thanks: 0 |z² - 2z -1| = 0 (Complex Variable)? z = x + iy. What is a better way to find that this is satisfied by the two points 1±√2 ? The way I'm doing it is turning into a huge mess of x's and y's. Here's how I'm doing it: z² = (x² - y²) + i(2xy) Letting w = z² - 2z -1 w = (x² -y² -2x -1) + i(2xy -2y) |w|² = (x² -y²-2x-1)² + (2xy - 2y)² = 0 This turned out to be a huge mess, and I couldn't even solve it. And sorry this is the closest category available for this question I could find.
 February 3rd, 2017, 11:59 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,860 Thanks: 966 $|z| = 0 \Rightarrow z=0$ $z^2 - 2z -1 = 0$ $(z-1)^2 -2 = 0$ $(z-1)^2 = 2$ $z-1 = \pm \sqrt{2}$ $z = \pm \sqrt{2} + 1$ $z = 1 \pm \sqrt{2}$ I really don't see any reason to do it any other way. It just adds complication. Thanks from topsquark, Country Boy and Robart

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### z²-2z=-1 i

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