Loci in Argand Diagrams (Cambridge (CIE) A Level Maths: Pure 3): Revision Note

Loci in Argand diagrams

How do I sketch $\mathbf{Re}\mathit{ }\mathit{z}=\mathit{k}$ or $\mathbf{Im}\mathit{ }\mathit{z}=\mathit{k}$ on an Argand diagram?

• All complex numbers, $z=x+\mathrm{i}y$, that satisfy the equation $\mathrm{Re}z=k$ lie on a vertical line with Cartesian equation $x=k$

  • Any complex number along this vertical line will have a real part of $k$

• All complex numbers, $z=x+\mathrm{i}y$, that satisfy the equation $\mathrm{Im} z=k$ lie on a horizontal line with Cartesian equation $y=k$

  • Any complex number along this horizontal line will have an imaginary part of $k$

• E.g. The loci $\mathrm{Re}z=4$ and $\mathrm{Im} z=-3$ are represented by the vertical line $x=4$ and the horizontal line $y=-3$  

Sketching the loci of  $\mathbf{Re}\mathit{z}\mathbf{=}\mathbf{4}$ and $\mathbf{Im}\mathit{z}\mathbf{=}\mathbf{-}\mathbf{3}$

How do I sketch $\left|\mathit{z}-\mathit{a}\right|=\mathit{k}$ on an Argand diagram?

• All complex numbers, $z$, that satisfy the equation $\left|z\right|=k$ lie on a circle of radius $k$ about the origin

  • E.g. the locus of $\left|z\right|=10$ is a circle of radius 10, centred at the origin, as every complex number on that circle has a modulus of 10

• For a given complex number, $a$, all complex numbers, $z$, that satisfy the equation $\left|z-a\right|=k$ lie on a circle of radius $k$ about the centre $a$

  • This is because $\left|z-a\right|$ represents the distance between complex numbers $z$ and $a$ 

  • E.g. the locus of $\left|z-\left(3+4\mathrm{i}\right)\right|=10$ is a circle of radius 10 about $(3+4\mathrm{i})$

• Many equations need to be adjusted algebraically into the correct $\left|z-a\right|$ form

  • E.g. to find the centre of the circle $\left|z-8\mathrm{i}+3\right|=12$, first rewrite it as $\left|z-\left(-3+8\mathrm{i}\right)\right|=12$, giving the centre as $-3+8\mathrm{i}$

  • E.g. to find the centre of the circle $\left|z+\mathrm{i}\right|=2$, first rewrite it as $\left|z-\left(-\mathrm{i}\right)\right|=2$ , giving the centre as $\left(-\mathrm{i}\right)$

  • Note that the centre of the circle $\left|z\right|=5$ is the origin (it can be thought of as $\left|z-0\right|=5$)

• In order to sketch correctly, check whether the origin lies outside, on or inside the circle

  • E.g. for the locus of $\left|z-\left(3+4\mathrm{i}\right)\right|=10,$ the distance from the centre of the circle, $3+4\mathrm{i}$, to the origin is 5 (by Pythagoras), which is less than the radius of 10; a sketch must therefore show the origin inside the circle

• By knowing the radius and centre of a circle, the Cartesian equation of the circle can be found

  • The circle $\left|z-\left(3+4\mathrm{i}\right)\right|=10$ has a radius of 10 and centre of $(3,4)$ in coordinates, so the equation of the circle is ${\left(x-3\right)}^2+{\left(y-4\right)}^2=100$

Sketching the loci of  $\left|z\right|\mathbf{=}\mathbf{10}$ and $\left|z-\left(3+4i\right)\right|\mathbf{=}\mathbf{10}$

How do I sketch $\left|\mathit{z}-\mathit{a}\right|=|\mathit{z}-\mathit{b}|$ on an Argand diagram?

• For two given complex numbers, $a$ and $b$, all complex numbers, $z$, that satisfy the equation $\left|z-a\right|=|z-b|$ lie on the perpendicular bisector of $a$ and $b$

  • This is because the distance from $z$ to $a$ must equal the distance from $z$ to $b$

    • a condition that is satisfied by all the complex numbers, $z$, on the perpendicular bisector of $a$ and $b$

  • E.g. the locus of $\left|z-3+2\mathrm{i}\right|=|z+8|$ can be rewritten as $\left|z-\left(3-2\mathrm{i}\right)\right|=\left|z-\left(-8\right)\right|$ which is the perpendicular bisector of the points $3-2\mathrm{i}$ and $-8$

• A sketch of the perpendicular bisector is sufficient, without finding its exact equation (though this could be found using coordinate geometry methods)

Sketching the loci of  $|\mathit{z}-i|=\left|\mathit{z}-3i\right|$and $|\mathit{z}-\left(3-2i\right)|=\left|\mathit{z}-\left(-8\right)\right|$

How do I sketch $\mathbf{arg}(\mathit{z}-\mathit{a})=\mathit{\alpha }$  on an Argand diagram?

• All complex numbers, $z$, that satisfy the equation $\arg z=\alpha$ lie on a half-line from the origin at an angle of $\alpha$ to the positive real axis

  • Although the half-line starts at the origin, the origin itself ($z=0$) does not satisfy the equation $\arg z=\alpha$  as $\arg 0$ is undefined (there is no angle at the origin)

  • To show the exclusion of $z=0$ from the locus of $\arg z=\alpha$, a small open circle at the origin is used

  • E.g. the locus of $\arg z=\frac{\mathrm{\pi }}{4}$ is a half-line of angle $\frac{\mathrm{\pi }}{4}$ to the positive real axis, starting from the origin, with an open circle at the origin

• For a given complex number, $a$, all complex numbers, $z$, that satisfy the equation $\arg (z-a)=\alpha$ lie on a half-line from the point $a$ at an angle of $\alpha$ to the positive real axis, with an open circle to show the exclusion of $z=a$

  • E.g. the locus of $\arg \left(z-1-5\mathrm{i}\right)=\frac{2\pi }{3}$ can be rewritten as $\arg \left(z-\left(1+5\mathrm{i}\right)\right)=\frac{2\pi }{3}$, which is a half-line of angle $\frac{2\pi }{3}$measured from the point $1+5\mathrm{i}$, with an open circle at $1+5\mathrm{i}$ to show its exclusion

• In some cases, the equation of the half-line can be found using a sketch to help

  • E.g. the locus of $\arg z=\frac{\pi }{4}$is the half-line $y=x$ for $x>0$  

  • E.g. the locus of $\arg \left(z-(8+5\mathrm{i}\right)=-\frac{\pi }{4}$ can be thought of, in coordinate geometry, as the half-line through $(8,5)$with gradient -1, giving $y=-x+13$  for  $x>8$

  • Whilst not examinable, the half-line equation for a more general angle, $\arg z=\alpha$, is $y=\left(\tan \alpha \right)x$ for $x>0$, as the $\mathrm{gradient}=\frac{\mathrm{opposite}}{\mathrm{adjacent}}=\tan \alpha$

Sketching the loci of  $\mathbf{arg}\mathit{z}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{4}}$and $\mathbf{arg}\mathbf{(}\mathit{z}\mathbf{-}\left(1+5i)\right)\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{3}}$