## Loci with complex numbers - modulus (singular)

## Loci with complex numbers - circles 1

Determine where the circle, generated by the equation $|z - (1+2i)| = 3$, intersects the real and imaginary axes.

The easiest way to solve this problem is to determine the Cartesian equation of the circle and solve using standard geometric methods

The Cartesian equation is given by $$ (x-1)^2 + (y-2)^2 = 3^2 $$ It follows that the real axis is interested when $y = 0$, this implies that $(x-1)^2 = 5$ and hence $x = 1\pm \sqrt{5}$. Similarly, the imaginary axis is intersected by $x = 0$. This implies $(y-2)^2 = 8$, hence $y = 2\pm \sqrt{8}$.

We have not completed the question yet however, as we have not expressed our answers as complex numbers!

Our final answers are: $$ z = 1+\sqrt{5},\;1 - \sqrt{5},\;(2 +\sqrt{8})i,\; (2-\sqrt{8})i $$

## Loci with complex numbers - circles 2

Let $P$ be the point on the circle closest to the origin. Let $C = (3 + 3i)$ be the centre of the circle. Then $P$ lies on the line segment $OC$.

We can also say $OP = \lambda OC$. Given that $|OC| = \sqrt{3^2 + 3^2} = 3\sqrt{2}$ and that $|PC| = 2$, it follows that $|OP| = 3\sqrt{2} - 2$

$$OP = \frac{|OP|}{|OC|}OC = \frac{3\sqrt{2} - 2}{3\sqrt{2}}(3+3i)$$