-Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$
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+Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$
+
+## Sketching complex graphs
+
+- **Straight line:** $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector) or $\operatorname{Arg}(z) = \theta$
+- **Circle:** $|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
+- **Locus:** $\operatorname{Arg}(z) \lt \theta$