# Curves and regions in the complex plane

In the Cartesian plane, we often graph sets of points which meet certain criteria or relations. We can also do this in the complex plane, making use of the different complex functions such as the modulus and the argument.

Example. The set of points $$z$$ which satisfy the relation $$\text{Re}(z)=1$$ is depicted in the plot, i.e. all the points which have a real part of 1.

We should interpret $$|z-w|$$ as the distance from point $$z$$ to point $$w$$.

Example. The set of points $$z$$ which satisfy the relation $$|z-(1+i)|=2$$ is depicted in the plot, i.e. all the points which are a distance of 2 from the point $$1+i$$.

Example. The set of points $$z$$ which satisfy the relation $$|z-2-i|=|z+i|$$ is depicted in the plot, i.e. all the points which are equidistant from $$2+i$$ and $$-i$$.

We should interpret $$\text{arg}(z-w)$$ as the argument of point $$z$$ from point $$w$$.

Example. The set of points $$z$$ which satisfy the relation $$-\frac{\pi}{4} \le \text{arg}(z-i) < \frac{\pi}{4}$$ is depicted in the plot, i.e. all the points which have an argument between $$-\frac{\pi}{4}$$ and $$\frac{\pi}{4}$$ from the point $$-i$$. Note that there is an open circle at $$-i$$ because the argument of 0 is undefined, and that the upper bound where $$\text{Arg}(z-i)=\frac{\pi}{4}$$ is a dashed line to indicate that it is not included in the region.

Sometimes there is no simple geometric interpretation, so in these situations, let $$z=x+yi$$ and find the Cartesian relation.

Example. The set of points $$z$$ which satisfy the relation $$|z|=\text{Im}(z)+1$$ is depicted in the plot. We let $$z=x+yi$$ to give us $$\sqrt{x^2+y^2}=y+1,$$ which simplifies to become the parabola $$y= \frac{1}{2}(x^2-1)$$.

## Problems

1. Sketch the points $$z$$ on the complex plane which satisfy the following relations:
1. $$\text{Im}(z)<-2$$
2. $$\text{Re}(z) \ge \text{Im}(z)$$
3. $$|z+i|\ge|z-3i|$$
4. $$|z-(1+i)|=|z-(-1-i)|$$
5. $$|z|=2$$
6. $$|z-3+i| \le 1$$
7. $$\text{arg}(z) = \frac{\pi}{3}$$
8. $$\text{Arg}(z-1) \ge \frac{\pi}{2}$$ (Recall that the principal argument lies between $$-\pi$$ and $$\pi$$)
9. $$z+\overline{z}=4$$
10. $$|z^2-(\overline{z})^2|<4$$
11. $$|z|\le 3$$ and $$0 \le \arg(z) < \frac{\pi}{6}$$
12. $$z \overline{z} \le 4$$ or $$-1 < \text{Re}(z) < 1$$
13. Hard. $$\text{Arg}\left( \frac{z-1}{z+1} \right) = \frac{\pi}{6}$$