getridofns
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- Aug 4, 2007
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- 2010
how to sketch
Re(z)=|z-1|?
Re(z)=|z-1|?
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complex locus (1 Viewer)
- Thread starter getridofns
- Start date
ssglain
Member
Let z = x + iy and solve algebraically. Scroll over the spoiler for the solution and explanation.
Re(z) = x, |z - 1| = √{(x - 1)² + (y - 0)²}
x = √{(x - 1)² + (y - 0)²}
x² = x² - 2x + 1 +y²
y² = 2x - 1
.: y² = 2(x - 1/2)
The condition Re(z)=|z-1| basically translates to "P(x, y) represents complex number z such that its distance from (1, 0) equals its perpendicular distance from the y-axis" This makes sense geometrically because the locus traces out a parabola with vertex (1/2, 0) and focal length 1/2 - i.e. focus (1, 0) and directrix x = 0. Any point on this locus is equidistant from the focus and the directrix.
x = √{(x - 1)² + (y - 0)²}
x² = x² - 2x + 1 +y²
y² = 2x - 1
.: y² = 2(x - 1/2)
The condition Re(z)=|z-1| basically translates to "P(x, y) represents complex number z such that its distance from (1, 0) equals its perpendicular distance from the y-axis" This makes sense geometrically because the locus traces out a parabola with vertex (1/2, 0) and focal length 1/2 - i.e. focus (1, 0) and directrix x = 0. Any point on this locus is equidistant from the focus and the directrix.
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