Loci in the Gauss plane

A locus is the set of all the points that satisfy a condition. In the Gauss plane the condition is an equation in $\displaystyle z$, and $\displaystyle |z_1-z_2|$ is the distance between two points. So a locus can be recognised in two ways:

• geometrically: read the condition as distances, and recognise a known shape;
• algebraically: substitute $\displaystyle {z}={x+yi}$, and get an equation in $\displaystyle x$ and $\displaystyle y$, as in analytic geometry.

In the coordinate systems of this lesson the $\displaystyle x$-axis is the real axis, and the $\displaystyle y$-axis is the imaginary axis. The locus is written there as an equation in $\displaystyle x$ and $\displaystyle y$.

$\displaystyle {|z-z_0|}={R}$: the distance of $\displaystyle z$ from the point $\displaystyle z_0$ is $\displaystyle R$ – a circle with centre $\displaystyle z_0$ and radius $\displaystyle R$. In the example below, $\displaystyle {|z-2+i|}={3}$:

• $\displaystyle {|z-(2-i)|}={3}$: a circle with centre $\displaystyle 2-i$ and radius 3.
• Algebraically: $\displaystyle {|(x-2)+(y+1)i|}={3}$, so $\displaystyle {(x-2)^2+(y+1)^2}={9}$.
• $\displaystyle |z-z_0|\le R$ is the disc – the circle and all the points inside it; $\displaystyle |z-z_0|>R$ – the points outside it.

$\displaystyle {|z-z_1|}={|z-z_2|}$: $\displaystyle z$ is at equal distances from $\displaystyle z_1$ and $\displaystyle z_2$ – the perpendicular bisector of the segment $\displaystyle z_1z_2$. In the example below, $\displaystyle {|z-1|}={|z-i|}$:

• The points at equal distances from $\displaystyle 1$ and $\displaystyle i$: the perpendicular bisector of the segment from $\displaystyle (1,0)$ to $\displaystyle (0,1)$.
• Algebraically: $\displaystyle {(x-1)^2+y^2}={x^2+(y-1)^2}$, and after cancelling $\displaystyle {-2x}={-2y}$, that is $\displaystyle {y}={x}$.

A condition on the real part or the imaginary part is an equation in $\displaystyle x$ and $\displaystyle y$ at once:

• $\displaystyle {\operatorname{Re}(z)}={3}$ is $\displaystyle {x}={3}$ – a line parallel to the imaginary axis.
• $\displaystyle {\operatorname{Im}(z)}={-1}$ is $\displaystyle {y}={-1}$.
• $\displaystyle {\operatorname{Im}(z)}={2\operatorname{Re}(z)+1}$ is $\displaystyle {y}={2x+1}$.
• $\displaystyle {z+\bar{z}}={4}$ is $\displaystyle {2x}={4}$, that is $\displaystyle {x}={2}$.

When the ratio of the distances from two points is constant, $\displaystyle m>0$ and $\displaystyle m\neq1$, the locus is a circle – Apollonius’ circle. In the example below, $\displaystyle {|z|}={2|z-3|}$:

• $\displaystyle {x^2+y^2}={4\left[(x-3)^2+y^2\right]}$.
• $\displaystyle {3x^2+3y^2-24x+36}={0}$, and dividing by 3: $\displaystyle {x^2+y^2-8x+12}={0}$.
• Completing the square: $\displaystyle {(x-4)^2+y^2}={4}$ – a circle with centre $\displaystyle 4$ and radius 2.

When $\displaystyle {m}={1}$ it is the perpendicular bisector.

Write the locus as an equation in $\displaystyle x$ and $\displaystyle y$ in a row of the board.
$\displaystyle {|z+1-3i|}={2}$

Write the locus as an equation in $\displaystyle x$ and $\displaystyle y$, and describe it in words.
$\displaystyle {|z+3i|}={|z-1|}$

Write the locus as an equation in $\displaystyle x$ and $\displaystyle y$, and find the centre and the radius.
$\displaystyle {|z+3|}={2|z|}$

Write the region as an inequality in $\displaystyle x$ and $\displaystyle y$.
$\displaystyle |z-1-i|\le2$

The board shows a circle.
Write its equation in $\displaystyle x$ and $\displaystyle y$, then write it in the form $\displaystyle {|z-z_0|}={R}$.

$\displaystyle {z\cdot\bar{z}+2\operatorname{Re}(z)}={8}$.
Write the locus as an equation in $\displaystyle x$ and $\displaystyle y$, and identify it.