7. Locus

L3CALC

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Locus (plural: loci) are a set of points that meet a certain condition.

For example (of a locus), when we take a circle, a circle has infinite amount of points that satisfy one condition:

The points must have a distance $r$ from the center of the circle.

Another example of locus is a line. A linear line meets one condition too:

The points must be on the the gradient of the line.

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In L3 Complex Numbers, locus questions come out at usually Excellence (sometimes Merit) level. In order to gain a high mark in locus questions is that we must follow the steps carefully.

For example, in 2022 Complex Numbers (in Q1e):

$$|z+i|=2|z-5i|$$

The following locus was given, where we were supposed to find the cartesian equation of the locus in the form: $(x-a{)}^2+(y-b{)}^2=k^2$

Steps Explained:

1. First thing we do is we must write $z$ in another way. Whenever we are representing in a ‘cartesian’ equation, we have to know the $x$ and $y$ of the equation.

   In Complex Numbers, we have learnt that $z$ can be expressed in the form $x+yi$ (or $a+bi$, but we use $x+yi$ for locus questions for easier working).

   We must write $z$ as $x+yi$ for the start:

   $$|x+yi+i|=2|x+yi-5i|$$

2. Next thing we do is we look at the modulus (the absolute value signs). They represent the distance from the origin, and that equation can be modelled by:

   $$r=\sqrt{x^2+y^2}$$

   In that equation, when we switch that to the Argand plane, $x$ becomes the real term, and $y$ becomes the imaginary terms. In the question, we have both real AND imaginary terms inside the modulus. Now, we should group the real terms, and group the imaginary terms using a bracket:

   $$|x+(y+1)i|=2|x+(y-5)i|$$

   Now, we can remove the absolute value signs by using the modulus formula:

   $$\sqrt{x^2+(y+1{)}^2}=2\sqrt{x^2+(y-5{)}^2}$$

3. Next thing we should do is get rid of the square root, which can be done by squaring both sides:

$$x^2+(y+1{)}^2=4(x^2+(y-5{)}^2)$$

4. Now, we expand and simplify:

   $$x^2+y^2+2y+1=4(x^2+y^2-10y+25)$$ $$x^2+y^2+2y+1=4x^2+4y^2-40y+100$$ $$-3x^2-3y^2+42y-99=0$$ $$x^2+y^2-14y+33=0$$ $$x^2+(y^2-14y+49)-49+33=0$$ $$x^2+(y-7{)}^2=16$$

5. Now, we have the cartesian equation of a circle, in the form requested by the question. Although this would get us the Excellence mark, it’s a good practice to state the center and radius of the locus (circle) we got: $center:(0,7)$ & $radius=4$

Steps Summary:

1. Turn $z$ into $x+yi$
2. Group real terms by real terms, and imaginary terms by imaginary terms
3. Use square root to represent the modulus
4. Expand & simplify
5. State cartesian equation of locus