## - Locus on an Argand Diagram

LOCUS ON THE ARGAND DIAGRAM

A locus is a set of points which satisfy a given condition.
THE CIRCLE
The equation $$|z-z_1|=r$$ where $$z$$ and $$z_1$$ are complex numbers of the form $$x+yi$$ and $$r$$ is a constant represents a circle with centre $$z_1$$ and radius $$r$$.
RECALL: The equation of a circle is $$(x-h)^2+(y-k)^2=r^2$$ where $$(h, k)$$ are the coordinates of the circle and $$r$$ is the radius of the circle.

LESSON 1
If the point $$P$$ in the complex plane corresponds to the complex number $$z$$, find the locus of $$P$$ in each of the following situations.
(i) $$|z|=3$$
(ii) $$|z-2|=4$$
(iii) $$|z+3-i|=2$$
SOLUTION
(i) $$|z|=3$$
• Write in the form $$|z-z_1|=r$$
$$|z-0|=3$$
The locus is therefore a circle with centre (0, 0) and radius 3.
This can be written algebraically as
$$x^2+y^2=9$$
Therefore, $$x^2+y^2=9$$ is the Cartesian equation of the locus.
(ii) $$|z-2|=4$$
Equation is written in the form $$|z-z_1|=r$$. Therefore, the locus is a circle with centre (2, 0) and radius 4.
Algebraically, $$(x-2)^2+y^2=4^2$$ is the equation of the locus.
(iii) $$|z+3-i|=2$$
• Write in the form $$|z-z_1|=r$$
$$|z-(-3+i)|=2$$
It can now be stated that the locus is a circle of radius 2 and centre (-3,1).
$$(x+3)^2+(y-1)^2=4$$
Let’s take a look at how these loci can be determined algebraically.
$$|z+3-i|=2$$
• Replace $$z$$ with $$x+yi$$
$$|x+yi+3-i|=2$$
• Group real and imaginary parts.
$$|x+3+(y-1)i|=2$$
• Square both sides to remove the modulus sign.
$$(x+3)^2+(y-1)^2=4$$

CARTESIAN EQUATIONS
LESSON 1
Determine the Cartesian equation of the locus of points satisfying the following conditions.
(i) $$2|z-3i|=|z|$$
(ii) $$\displaystyle |{z-1\over z+2}|=\sqrt{3}$$
SOLUTION
(i) $$2|z-3i|=|z|$$
• Replace $$z$$ with $$x+yi$$
$$2|x+yi-3i|=|x+yi|$$
• Group real and imaginary parts together.
$$2|x+(y-3)i|=|x+yi|$$
• Square both sides to remove moduli signs.
$$4[x^2+(y-3)^2 ]=x^2+y^2$$
• Simplify
$$4x^2+4y^2-24y+36=x^2+y^2$$
$$3x^2+3y^2-24y+36=0$$
$$x^2+y^2-8y+12=0$$
$$x^2+(y-4)^2=4$$

(ii) $$\displaystyle |{z-1\over z+2}|=\sqrt{3}$$
• Use the fact that $$\displaystyle |{z_1\over z_2}|={|z_1|\over|z_2|}$$
$$\displaystyle |{z-1 \over z+2}|={|z-1|\over|z+2|}=\sqrt{3}$$
$$|z-1|=\sqrt{3} |z+2|$$
• Replace $$z$$ with $$x+yi$$
$$|x+yi-1|=\sqrt{3} |x+yi+2|$$
• Group real and imaginary parts
$$|x-1+yi|=\sqrt{3} |x+2+yi|$$
• Square both sides
$$(x-1)^2+y^2=3[(x+2)^2+y^2 ]$$
• Simplify
$$x^2-2x+1+y^2=3x^2+12x+12+3y^2$$
$$2x^2+2y^2+14x+11=0$$
$$\displaystyle x^2+7x+y^2+{11\over2}=0$$
$$\displaystyle \left(x+{7\over2}\right)^2+y^2={27\over4}$$

LESSON 3
Sketch the locus of the point $$P(x, y)$$ representing the complex number $$z=x+yi$$, given that $$|z-3i|=|z+2+5i|$$. Write down the Cartesian equation of the locus.
SOLUTION
$$|z-3i|=|z+2+5i|$$
• Rewrite both sides in the form $$|z-z_1|$$
$$|z-3i|=|z-(-2-5i)|$$
• We can state that the distance between $$P(x, y)$$ and the complex number represented by the point (0, 3) on the Argand diagram is equal to the distance between $$P(x, y)$$ and the complex number represented by the point (-2, -5) on the Argand diagram. Consequently, we can deduce that the locus is the perpendicular bisector of these two points.

• To determine the Cartesian equation we can proceed in two ways.
METHOD 1:
• Determine the mid – point of the line segment joining (0,3) and (-2, -5)
Mid-point$$\displaystyle =\left({x_2+x_1\over2}, {y_2+y_1\over2}\right)=\left({0+(-2)\over2}, {3+(-5)\over2}\right)=(-1, -1)$$
• Determine the gradient of the line segment joining (0, 3) and (-2, -5).
$$\displaystyle m={y_2-y_1\over x_2-x_1}={3-(-5)\over 0-(-2)}=4$$
$$\displaystyle m_⊥=-{1\over4}$$
• Determine the equation of the perpendicular bisector y=mx+c.
$$y=mx+c$$ using $$\displaystyle m_⊥=-{1\over4}$$ and (-1, -1)
$$\displaystyle -1=-{1\over4}(-1)+c$$
$$\displaystyle -{5\over4}=c$$
$$\displaystyle y=-{1\over4} x-{5\over4}$$
METHOD 2:
$$|z-3i|=|z-(-2-5i)|$$
Let $$z=x+yi$$
$$|x+yi-(0+3i)|=|x+yi-(-2-5i)|$$
• Group like terms together
$$|x+(y-3)|=|(x+2)+(y+5)|$$
• Square both sides
$$x^2+(y+3)^2=(x+2)^2+(y+5)^2$$
• Simplify
$$x^2+y^2-6y+9=x^2+4x+4+y^2+10y+25$$
$$6y+4x+20=0$$
$$4y+x+5=0$$
$$\displaystyle y=-{1\over4} x-{5\over4}$$

HALF LINES
The equation $$\arg⁡(z-z_1)=θ$$ produces a half line starting at the point represented by $$z_1$$ on the Argand diagram and which makes an angle of $$θ$$ with the positive $$x$$ – axis.
NB: $$z_1$$ is NOT included on the half line.

LESSON 1
Describe and sketch the locus of the points satisfying the following conditions.
(a) $$\displaystyle \arg⁡(z-3)={π\over4}$$
(b) $$\displaystyle \arg⁡(z+3-2i)={π\over3}$$
SOLUTION
(i) $$\displaystyle \arg⁡(z-3)={π\over4}$$
This is a half – line which starts at (3,0) (not including (3, 0)) and makes an angle of $$\displaystyle {π\over4}$$ with the positive $$x$$ – axis.
Additionally, the Cartesian equation of this line can be determined as follows:
$\arg⁡(z-3)={π\over4}$
Let $$z=x+yi$$
$$\displaystyle \arg⁡[(x+yi)-(3-0i)]={π\over4}$$
• Group real and imaginary parts together.
$$\displaystyle \arg⁡[(x-3)+yi]={π\over4}$$
$$\displaystyle \tan^{-1}\left({y\over x-3}\right)={π\over4}$$
$$\displaystyle {y\over x-3}=\tan⁡\left({π\over4}\right)=1$$
$$y=x-3; x>3$$

(ii) $$\displaystyle \arg⁡(z+3-2i)={π\over3}$$
• Rewrite in the form $$\arg⁡(z-z_1)=θ$$
$$\displaystyle \arg⁡(z-(-3+2i))={π\over3}$$
This is a half line starting at the point (-3, 2) exclusive, which makes an angle of $$\displaystyle {π\over3}$$ with the positive $$x$$ – axis.
The algebraic determination of this line would be as follows:
$$\displaystyle \arg⁡[(x+yi)-(-3+2i)]={π\over3}$$
$$\displaystyle \arg⁡[(x+3)+(y-2)i]={π\over3}$$
$$\displaystyle {y-2\over x+3}=\tan\left({π\over3}\right)=\sqrt{3}$$
$$y-2=\sqrt{3} x+3\sqrt{3}$$
$$y=\sqrt{3} x+2+3\sqrt{3}; x>-3$$

INEQUALITIES
LESSON 1
Shade on an Argand diagram the region in which $$|z-2i|≤1$$
SOLUTION
• Determine the locus generated by
$$|z-2i|=1$$
$$|z-(0+2i)|=1$$
Circle with centre (0, 2) and radius 1.
• Shade inside of this circle since we want
$$|z-2i|≤1$$

LESSON 2
(a) Sketch on one Argand diagram:
(i) the locus of points satisfying
$|z-i|=|z-2|$
(ii)the locus of points satisfying
$\arg⁡(z-i)={π\over4}$
(b) Shade on your diagram the region in which $$|z-i|≤|z-2|$$ and $$\displaystyle –{π\over2}≤\arg ⁡(z-i)≤{π\over4}$$
SOLUTION
(a) (i) $$|z-i|=|z-2|$$
$$|(x+yi)-(0+i)|=|(x+yi)-(2+0i)|$$
This is the perpendicular bisector of the line segment joining the points (0,1) and (2,0)
(ii) $$\displaystyle \arg⁡(z-i)={π\over4}$$
$$\displaystyle \arg⁡[(x+yi)-(0+i)]={π\over4}$$
Half – line starting at (0, 1), exclusive, making an angle of $$\displaystyle {π\over4}$$ with the positive $$x$$-axis.
(b)The distance between $$z-2$$ is greater than the distance between $$z-i$$ above the first line. The inequality includes equal to so we also include the line. Since $$\displaystyle -{π\over2}≤\arg⁡(z-i)≤{π\over4}$$, the required region starts at the half line and rotates clockwise to the $$y$$ – axis. The overlap of these two regions is shown below.

LESSON 3
(a) Sketch on an Argand diagram the locus of points satisfying the equation
$|z-6i|=3$
(b) It is given that z satisfies the equation $$|z-6i|=3$$.
(i) Write down the greatest possible value of $$|z|$$.
(ii) Find the greatest possible value of $$\arg ⁡z$$, giving your answer in the form $$pπ$$, where $$-1<p≤1$$.
SOLUTION
(a) Circle with centre (0,6) and radius 3.
(b) (i) 9 is the largest possible value of $$|z|$$. This would occur when $$z=9i$$.
(ii) When the line segment connecting the origin and $$z$$ is tangential to the circle $$\arg ⁡z$$ has its greatest value. The point of tangency, the point (0, 6) and the origin form a right-angled triangle (a radius and a tangent form a right angle). The required angle comprises a right angle (between the positive $$x$$ – axis and the positive $$y$$ – axis) plus the angle formed by the positive $$y$$ – axis and the tangent.
NB: The radius of the circle is 3 and the distance from the origin to the point (0, 6) is 6.
$\arg⁡(z)={π\over2}+\sin^{-1}\left({3\over6}\right)={2π\over3}$

POINTS OF INTERSECTION
LESSON 1
(a) On the same Argand diagram, sketch the loci of points satisfying:
(i) $$|z+3+i|=5$$
(ii) $$\displaystyle \arg⁡(z+3)=-{3π\over4}$$
(b) (i) From your sketch, explain why there is only one complex number satisfying both equations.
(ii) Verify that this complex number is $$-7-4i$$
SOLUTION
(a) (i) $$|z+3+i|=5$$
$$|z-(-3-i)|=5$$
Circle with centre (-3, -1) and radius 5
(ii) $$\displaystyle \arg⁡(z+3)=-{3π\over4}$$
$$\displaystyle \arg⁡(z-(-3))=-{3π\over4}$$
Half – line, starting at (-3, 0), exclusive, making an angle of $$\displaystyle -{3π\over4}$$ with the positive real axis.
(b) (i) There is only one complex number satisfying both equations since there is only one point of intersection due to the half-line which starts within the circle.
(ii) If $$-7-3i$$ is the point of intersection it must satisfy both conditions.
$$|-7-4i+3+i|=|-4-3i|$$
$$=\sqrt{(-4)^2+(-3)^2}$$
$$=5$$
$$\arg⁡(-7-4i+3)=\arg⁡(-4-4i)$$
$$\displaystyle =-π+\tan^{-1}⁡\left({4 \over 4}\right)$$
$$\displaystyle =-{3π\over4}$$