Have I graphed this complex equation correctly?

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November 19th 2011, 10:32 PM
terrorsquid

Have I graphed this complex equation correctly?

Find and graph all $z \in \mathbb{C}$ that satisfy the equation $|z-2i|=|z+1|$ .

So, I am finding the values where the distance from z to 2i is the same as the distance from z to -1 right?

My solution:

$|(x+yi)-2i|=|(x+yi)+1|$

$\sqrt{x^2+(y-2)^2}=\sqrt{(x+1)^2+y^2}$

$x^2+(y-2)^2}=(x+1)^2+y^2$

$x^2+y^2-4y+4=x^2+y^2+2x+1$

$y=-\frac{1}{2}x+\frac{3}{4}$

Is there anything more to it than that; now that I have the line?

Thanks.
November 20th 2011, 01:37 AM
Prove It

Re: Have I graphed this complex equation correctly?

Quote:

Originally Posted by terrorsquid

Find and graph all $z \in \mathbb{C}$ that satisfy the equation $|z-2i|=|z+1|$ .

So, I am finding the values where the distance from z to 2i is the same as the distance from z to -1 right?

My solution:

$|(x+yi)-2i|=|(x+yi)+1|$

$\sqrt{x^2+(y-2)^2}=\sqrt{(x+1)^2+y^2}$

$x^2+(y-2)^2}=(x+1)^2+y^2$

$x^2+y^2-4y+4=x^2+y^2+2x+1$

$y=-\frac{1}{2}x+\frac{3}{4}$

Is there anything more to it than that; now that I have the line?

Thanks.

Yes, now you have to graph the line...
November 20th 2011, 03:09 AM
HallsofIvy

Re: Have I graphed this complex equation correctly?

As you say, this is the set of all points that are equidistant from 2i and -1 or, in terms of x and y, from (0, 2) and (-1, 0). It is well known, geometrically, that the set of all points equidistant from two given points is the "perpendicular bisector" of the line segment between the given points.

The slope of the line from (0, 2) to (-1, 0) is (2- 0)/(0-(-1)= 2 so the slope of the perpendicular bisector is -1/2. The midpoint of (0, 2) to (-1, 0) is ((0- 1)/2, (2+ 0)/2)= (-1/2, 1) so you want the line through (-1/2, 1) with slope -1/2. That is, of course, y- 1= (-1/2)(x+ 1/2) or y= (-1/2)x+ 3/4, just as you give.