Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Loci

In H2 Mathematics, we learnt several standard loci. Let us look at an example of a non-standard one and understand its problem solving approach.

Question: Given that \text{arg}(z + \text{i}) - \text{arg}(z - \text{i}) = \frac{\pi}{4}, show that the locus of the point representing z is a circle.

Solution: First, we can rewrite the equation as \text{arg}\left( \frac{z + \text{i}}{z - \text{i}} \right) = \frac{\pi}{4}.

Next, we express \frac{z + \text{i}}{z - \text{i}} in the form a + b\text{i}, by letting z = x + y\text{i}.

So \frac{z + \text{i}}{z - \text{i}} = \frac{x + y\text{i} + \text{i}}{x + y\text{i} - \text{i}} which can be simplified to \frac{x^2 + y^2 - 1}{x^2 + (y - 1)^2} + \frac{2x}{x^2 + (y - 1)^2}\text{i}, by the usual method you have been taught in H2 mathematics (try it!).

Now, any line whose gradient is 1 makes an angle of \frac{\pi}{4} with respect to the horizontal. This is an important point to remember always. 

So we have \frac{2x}{x^2 + (y - 1)^2} = \frac{x^2 + y^2 - 1}{x^2 + (y - 1)^2}, which simplifies to 2x = x^2 + y^2 - 1.

Upon completing the square, we arrive at (x - 1)^2 + y^2 = 2, which is a circle whose centre is at (1, 0) and having a radius of \sqrt{2} units.