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Mathematics · Ch 2 — Complex Numbers

Properties of Modulus of a Complex Number

2.5.1

Properties of Modulus of a Complex Number

Eight properties of the modulus (for complex numbers z,z1,z2z,z_1,z_2 and integer nn):

  1. ∣z∣=∣z‾∣|z|=|\overline z|
  2. ∣z1+z2∣≤∣z1∣+∣z2∣|z_1+z_2|\le|z_1|+|z_2| (Triangle Inequality)
  3. ∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|
  4. ∣z1z2∣=∣z1∣∣z2∣, z2≠0\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|},\ z_2\ne0
  5. ∣z1−z2∣≥∣∣z1∣−∣z2∣∣, z2≠0|z_1-z_2|\ge\big||z_1|-|z_2|\big|,\ z_2\ne0
  6. ∣zn∣=∣z∣n|z^n|=|z|^n
  7. Re⁡(z)≤∣z∣\operatorname{Re}(z)\le|z|
  8. Im⁡(z)≤∣z∣\operatorname{Im}(z)\le|z|

Proof — Triangle Inequality. Using ∣w∣2=ww‾|w|^2=w\overline w:

∣z1+z2∣2=(z1+z2)(z1+z2)‾=(z1+z2)(z1‾+z2‾)=z1z1‾+z1z2‾+z1‾z2+z2z2‾=∣z1∣2+2Re⁡(z1z2‾)+∣z2∣2|z_1+z_2|^2=(z_1+z_2)\overline{(z_1+z_2)}=(z_1+z_2)(\overline{z_1}+\overline{z_2})=z_1\overline{z_1}+z_1\overline{z_2}+\overline{z_1}z_2+z_2\overline{z_2}=|z_1|^2+2\operatorname{Re}(z_1\overline{z_2})+|z_2|^2

(since z1z2‾+z1‾z2=z1z2‾+z1z2‾‾=2Re⁡(z1z2‾)z_1\overline{z_2}+\overline{z_1}z_2=z_1\overline{z_2}+\overline{z_1\overline{z_2}}=2\operatorname{Re}(z_1\overline{z_2})). By property 7, Re⁡(z1z2‾)≤∣z1z2‾∣=∣z1∣∣z2‾∣=∣z1∣∣z2∣\operatorname{Re}(z_1\overline{z_2})\le|z_1\overline{z_2}|=|z_1||\overline{z_2}|=|z_1||z_2|, so

∣z1+z2∣2≤∣z1∣2+2∣z1∣∣z2∣+∣z2∣2=(∣z1∣+∣z2∣)2⟹∣z1+z2∣≤∣z1∣+∣z2∣.|z_1+z_2|^2\le|z_1|^2+2|z_1||z_2|+|z_2|^2=(|z_1|+|z_2|)^2\quad\Longrightarrow\quad |z_1+z_2|\le|z_1|+|z_2|.

It generalises to any finite number of terms: ∣z1+z2+⋯+zn∣≤∣z1∣+∣z2∣+⋯+∣zn∣|z_1+z_2+\cdots+z_n|\le|z_1|+|z_2|+\cdots+|z_n|.

Geometric interpretation. In the triangle with vertices O,z1,z1+z2O,z_1,z_1+z_2, the side corresponding to the vector z1+z2z_1+z_2 cannot be longer than the sum of the lengths of the other two sides — exactly the ordinary triangle inequality from geometry, which is where the property gets its name.

Distance property. If z1=x1+iy1z_1=x_1+iy_1 and z2=x2+iy2z_2=x_2+iy_2, then

z1−z2=(x1−x2)+i(y1−y2)⟹∣z1−z2∣=(x1−x2)2+(y1−y2)2.z_1-z_2=(x_1-x_2)+i(y_1-y_2)\quad\Longrightarrow\quad |z_1-z_2|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.

So ∣z1−z2∣|z_1-z_2| is exactly the ordinary distance between the points z1z_1 and z2z_2 in the Argand plane. Considering O,z1,z2O,z_1,z_2 as a triangle and applying the triangle inequality to its sides gives two useful bounds used repeatedly to estimate ∣z∣|z| from a given constraint: ∣z1+z2∣≤∣z1∣+∣z2∣|z_1+z_2|\le|z_1|+|z_2| and ∣z1−z2∣≥∣∣z1∣−∣z2∣∣|z_1-z_2|\ge\big||z_1|-|z_2|\big|. …

Figure 2.17Triangle inequality $|z_1+z_2|\le|z_1|+|z_2|$ shown on the addition parallelogram
Fig. 2.17 — Triangle inequality $|z_1+z_2|\le|z_1|+|z_2|$ shown on the addition parallelogram

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. Triangle inequality ∣z1+z2∣≤∣z1∣+∣z2∣|z_1+z_2|\le|z_1|+|z_2| shown on the addition parallelogr …

Figure 2.18Distance $|z_1-z_2|$ between two points as the third side of triangle $O z_1 z_2$
Fig. 2.18 — Distance $|z_1-z_2|$ between two points as the third side of triangle $O z_1 z_2$

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. Distance ∣z1−z2∣|z_1-z_2| between two points as the third side of triangle $O z_1 z_ …

Figure 2.19Example 2.11: the points $i,\,-2+i,\,3$ and their distances from the origin
Fig. 2.19 — Example 2.11: the points $i,\,-2+i,\,3$ and their distances 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. Example 2.11: the points i, −2+i, 3i,\,-2+i,\,3 and their distances from the origi …

Figure 2.20Example 2.13: for $|z|=2$, the value $|z+3+4i|$ lies between $3$ and $7$ (circle of radius 2 centred at $(-3,-4)$)
Fig. 2.20 — Example 2.13: for $|z|=2$, the value $|z+3+4i|$ lies between $3$ and $7$ (circle of radius 2 centred at $(-3,-4)$)

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. Example 2.13: for ∣z∣=2|z|=2, the value ∣z+3+4i∣|z+3+4i| lies between 33 and 77 (circle of radius 2 centred at $(- …

Figure 2.21Fig 2.21: the points $1,\ -\frac{1}{2}+i\frac{\sqrt{3}}{2},\ -\frac{1}{2}-i\frac{\sqrt{3}}{2}$ as vertices of an equilateral triangle in the Argand plane
Fig. 2.21 — Fig 2.21: the points $1,\ -\frac{1}{2}+i\frac{\sqrt{3}}{2},\ -\frac{1}{2}-i\frac{\sqrt{3}}{2}$ as vertices of an equilateral triangle in the Argand plane

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 2.21: the points 1, −12+i32, −12−i321,\ -\frac{1}{2}+i\frac{\sqrt{3}}{2},\ -\frac{1}{2}-i\frac{\sqrt{3}}{2} as vertices of an equilateral triangle in the …