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February 3rd, 2017, 11:45 AM   #1
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|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.
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February 3rd, 2017, 11:59 AM   #2
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$|z| = 0 \Rightarrow z=0$

$z^2 - 2z -1 = 0$

$(z-1)^2 -2 = 0$

$(z-1)^2 = 2$

$z-1 = \pm \sqrt{2}$

$z = \pm \sqrt{2} + 1$

$z = 1 \pm \sqrt{2}$

I really don't see any reason to do it any other way. It just adds complication.
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