4DOtherCurves

Other Simple Curves in the Complex Plane

Different complex equations can produce unexpectedly familiar curves.

Often it is easiest to find the cartesian equivalent and sketch that.

This is done by substituting **z = x + yi**.

Example

math \text{Sketch } \left| z+2 \right| = \left| z+i \right| math


 * Solution:**

Substitute z = x + yi

math \\ . \qquad \left| x+yi+2 \right| = \left| x+yi+i \right| \\. \\ . \qquad \left| (x+2)+yi \right| = \left| x + (y+1)i \right| math

Find modulus of both sides

math . \qquad \sqrt{(x+2)^2+y^2} = \sqrt{x^2+(y+1)^2} math

Square both sides and simplify

math \\ . \qquad (x+2)^2+y^2=x^2+(y+1)^2 \\. \\ . \qquad x^2+4x+4+y^2 = x^2+y^2+2y+1 \\. \\ . \qquad 4x+4 = 2y+1 \\. \\ . \qquad 2y = 4x+3 \\. \\ . \qquad y = 2x + \frac{3}{2} math

So **|z + 2| = |z + i|** forms a straight line.

Notice that the line bisects and is perpendicular to a line segment joining –**2 + 0i** and 0 – **i**.

We can interpret the line
 * |z – (–2 + 0i)| = |z – (0 – i)|

as the set of points, P,
 * where the distance from –**2 + 0i** to P
 * is the equal to the distance from 0 – **i** to P

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