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if $Z=X+iy$ then determine the locus of the equation $\left | 2Z-1 \right | = \left | Z-2 \right |$.I can tell that it a circle equation and it is $x^2 + y^2 = 1$.There are a lot of equation in my book such as $\left | Z-8 \right | +\left | Z+8 \right |=20$,$\left | Z-2 \right | = \left | Z-3i \right |$,$\left | 2Z+3 \right |= \left | Z+6 \right |$.Every time I have to do a long calculation.
Is there any short way to find out that if the given equation is circle,ellipse, parabola, hyperbola or straight line.This is needed for my MCQ exam.

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$|2z-1| = |z-2|$:

Rearrange to $|z-\frac12| = \frac12|z-2|$: distance from $\frac12$ is half distance from $2$.

So it's a circle (distance from one point is a constant multiple $-$ not $0$ or $1$ $-$ of distance from another point). You know two points on the diameter: $1$ and $-1$. So you know the centre and radius.

$|z-8|+|z+8|=20$:

Distance from one point + distance from another point is constant: an ellipse with the two points as foci.

$|z-2| = |z-3i|$:

Distance from one point = distance from another point: this is the perpendicular bisector of the two points.

$|2z+3| = |z+6|$:

Rearrange to $|z+\frac32| = \frac12|z+6|$ and it's the same case as the first equation.

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Some fundamental equations:

$|z - z_0| = \rho$: circle with center $K(z_0)$ and radius $\rho$.

$|z - z_1| = |z - z_2|$: perpendicular bisector of the line segment with end points $K(z_1)$ and $K(z_2)$.

$|z-z_1| + |z-z_2| = 2a, \, a>0 \text{ and } |z_1-z_2|<2a $: ellipse with foci $E(z_1),E'(z_2)$ and constant sum $2a$ (recall the definition of the ellipse.)

$|z - a| = |Re(z) + a|,\, a\in \mathbb R \iff y^2 = 4ax: \text { parabola.}$

$|z - a\cdot i| =|Im(z) + a|,\, a\in \mathbb R \iff x^2 = 4ay: \text{ parabola.}$

$\big||z-z_1| - |z-z_2|\big|=2a,\, a>0 \text{ and } 2a<|z_1 - z_2|$: hyperbola with foci $E(z_1),E'(z_2)$ and constant absolute difference $2a$ (recall the definition of the hyperbola).


Note: In general, $|z-z_0|$ expresses the distance between the images $M(z)$ and $M(z_0)$.

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