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Mathematics · Ch 2 — Complex Numbers

Geometry and Locus of Complex Numbers

2.6

Geometry and Locus of Complex Numbers

This section studies the geometric interpretation of a complex number zz in the complex plane, and how to convert a condition on zz (given in terms of z,z‾,∣z∣z,\overline z,|z| or arg⁡z\arg z) into an ordinary Cartesian equation in x,yx,y — the equation of the locus traced out by zz.

Definition (circle). A circle is the locus of a point that moves in a plane such that its distance from a fixed point in that plane is always a constant. The fixed point is the centre, and the constant distance is the radius.

Complex form of the equation of a circle. Since ∣z−z0∣|z-z_0| is the distance from zz to the fixed point z0z_0, the locus of zz satisfying

∣z−z0∣=r(z0 fixed,r>0)|z-z_0|=r\qquad(z_0\text{ fixed}, r>0)

consists of all points at distance rr from z0z_0 — exactly a circle with centre z0z_0 and radius rr. Correspondingly:

  • ∣z−z0∣<r|z-z_0|<r represents the points interior to the circle;
  • ∣z−z0∣>r|z-z_0|>r represents the points exterior to the circle.

For instance, ∣z∣=r|z|=r (i.e. z0=0z_0=0) gives x2+y2=r\sqrt{x^2+y^2}=r, i.e. x2+y2=r2x^2+y^2=r^2: a circle centred at the origin, radius rr. An equation such as ∣αz−β∣=γ|\alpha z-\beta|=\gamma (α≠0\alpha\ne0) is first divided through by ∣α∣|\alpha| and rewritten as ∣z−βα∣=γ∣α∣\left|z-\dfrac\beta\alpha\right|=\dfrac\gamma{|\alpha|}, from which the centre βα\dfrac\beta\alpha and radius γ∣α∣\dfrac\gamma{|\alpha|} can be read off directly.

General loci. More general conditions on zz also trace out recognisable curves once translated to Cartesian form:

  • ∣z−a∣=∣z−b∣|z-a|=|z-b| (equidistant from two fixed points a,ba,b) always gives the perpendicular bisector of the segment joining aa and bb — a straight line, not a circle.
  • Conditions phrased with Re⁡(⋯ )\operatorname{Re}(\cdots), Im⁡(⋯ )\operatorname{Im}(\cdots) or z‾\overline z (e.g. [Re⁡(iz)]2=3[\operatorname{Re}(iz)]^2=3, Im⁡[(1−i)z+1]=0\operatorname{Im}[(1-i)z+1]=0, z‾=z−1\overline z=z^{-1}) become ordinary Cartesian equations once z=x+iyz=x+iy is substituted and the real/imaginary parts are separated — the result may be a line, a circle, or another simple curve.
  • An argument condition, e.g. arg⁡ ⁣(z−iz+2)=π4\arg\!\left(\dfrac{z-i}{z+2}\right)=\dfrac\pi4, fixes the angle subtended by the segment joining two fixed points, and typically traces an arc of a circle. …
Figure 2.22Fig 2.22: for $z=3+2i$, the points $A=z$, $B=z+iz$, $C=iz$ form an isosceles right triangle; dashed vectors show $z$, $iz$, $z+iz$ from the origin
Fig. 2.22 — Fig 2.22: for $z=3+2i$, the points $A=z$, $B=z+iz$, $C=iz$ form an isosceles right triangle; dashed vectors show $z$, $iz$, $z+iz$ from the origin

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 2.22: for z=3+2iz=3+2i, the points A=zA=z, B=z+izB=z+iz, C=izC=iz form an isosceles right triangle; dashed vectors show zz, iziz, z+izz+iz fro …

Figure 2.23Fig 2.23: the circle $|z-z_0|=r$ — locus of points $z$ at a fixed distance $r$ from the centre $z_0$
Fig. 2.23 — Fig 2.23: the circle $|z-z_0|=r$ — locus of points $z$ at a fixed distance $r$ from the centre $z_0$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 2.23: the circle ∣z−z0∣=r|z-z_0|=r — locus of points zz at a fixed distance rr from the centre …

Figure 2.24Fig 2.24: the circle $|3z-5+i|=4$, i.e. $|z-z_0|=\frac{4}{3}$, with centre $z_0=\left(\frac{5}{3},-\frac{1}{3}\right)$ and radius $\frac{4}{3}$
Fig. 2.24 — Fig 2.24: the circle $|3z-5+i|=4$, i.e. $|z-z_0|=\frac{4}{3}$, with centre $z_0=\left(\frac{5}{3},-\frac{1}{3}\right)$ and radius $\frac{4}{3}$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 2.24: the circle ∣3z−5+i∣=4|3z-5+i|=4, i.e. ∣z−z0∣=43|z-z_0|=\frac{4}{3}, with centre z0=(53,−13)z_0=\left(\frac{5}{3},-\frac{1}{3}\right) and radius $ …

Figure 2.25Fig 2.25: $|z+2-i|<2$ — the interior of the circle of centre $z_0=-2+i$ and radius $r=2$
Fig. 2.25 — Fig 2.25: $|z+2-i|<2$ — the interior of the circle of centre $z_0=-2+i$ and radius $r=2$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig 2.25: ∣z+2−i∣<2|z+2-i|<2 — the interior of the circle of centre z0=−2+iz_0=-2+i and radius $r …