e.g. {${z:|z|=4,z\in \mathbb{C}}$} (a circle at the centre with radius 4) ($z\in \mathbb{C}$)
What is {$z:|z|\leq{4}$}?
Is the same as before, but we shade the inside!! (conventionally with diagonal lines)
Alternatively, {$z:|z|<4$} Has a dotted line boundary and a shaded inside.
What is {$z:arg \ z=\frac{\pi}{6}$}
It makes a ray (line) going off to infinity and makes an angle of $\frac{\pi}{6}$ with the positive Real axis.
Now consider {$z:|z+8|=|z-4i|$}
Think of coordinate geometry, i.e. represent this things as x and y coordinates, then re-arrange until it is in a suitable plottable form.
$$\begin{align} \text{Let }&z=x+yi \\ \text{Then }&|x+yi+8|=|x+yi-4i| \\ &\sqrt{ (x+8)^{2}+y^{2} }=\sqrt{ x^{2}+(y-4)^{2} } \\ &(x+8)^{2}+y^{2}=x^{2}+(y-4)^{2} \\ &x^{2}+16x+64+y^{2}=x^{2}+y^{2}-8y+16 \\ &\therefore y=-2x-6 \\ &\text{Hence, it is a straight line with a } \\ &\text{y-intercept of -6 and gradient of -2}. \end{align} $$Alternatively,
$$\text{|z-a| is the distance between the complex number }z \text{ and }a. $$Hence,
$$\begin{align} \{ z:|z+8|=|z-4i\}\implies\{z:|z--8|=|z-4i|\} \\ \text{Hence, it is the points equidistant from -8 and 4i.} \\ \text{It will also be perpendicular to the line} \\ \text{between the two points -8 and 4i.} \end{align} $$Hence,
$$\begin{align} &Grad(\text{-8 to 4i})=\frac{4-0}{0--8}=\frac{1}{2} \\ &\therefore Grad(\text{locus})\times Grad(\text{-8 to 4i})=-1 \implies Grad(\text{locus})=-2 \\ &M.P=\left( -\frac{-8+0}{2}, \frac{0+4}{2} \right)=(-4,2) \\ &\therefore y-2=-2(x-(-4)) \implies y=-2x-6 \\ &\text{As we discovered from above.} \end{align} $$
