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February 3rd, 2017, 12:45 PM  #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) w² = (x² y²2x1)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, 12:59 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 817 
$z = 0 \Rightarrow z=0$ $z^2  2z 1 = 0$ $(z1)^2 2 = 0$ $(z1)^2 = 2$ $z1 = \pm \sqrt{2}$ $z = \pm \sqrt{2} + 1$ $z = 1 \pm \sqrt{2}$ I really don't see any reason to do it any other way. It just adds complication. 

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