- 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).
\(\displaystyle m={y_2-y_1\over x_2-x_1}={3-(-5)\over 0-(-2)}=4\)
  • Determine the perpendicular gradient
\(\displaystyle m_⊥=-{1\over4}\)
  • 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.
\(|-7-4i+3+i|=|-4-3i|\)
\(=\sqrt{(-4)^2+(-3)^2}\)
\(=5\)
\(\arg⁡(-7-4i+3)=\arg⁡(-4-4i)\)
\(\displaystyle =-π+\tan^{-1}⁡\left({4 \over 4}\right)\)
\(\displaystyle =-{3π\over4}\)