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Modulus and conjugate

Two of the most important operations on complex numbers are the modulus (size) and the conjugate (reflection in the real axis). They satisfy a dozen clean identities, and together they let us divide complex numbers and prove inequalities.

Definitions

Let z=a+biz = a + bi with a,b∈Ra, b \in \mathbb{R}.

The modulus (or absolute value) of zz is ∣z∣=a2+b2.|z| = \sqrt{a^2 + b^2}. It is the distance of zz from the origin in the Argand plane. For real zz, this agrees with the ordinary ∣x∣|x| , so the notation is unambiguous.

The conjugate of zz is zˉ=a−bi.\bar{z} = a - bi. It is the reflection of zz across the real axis.

The master identity

z zˉ=(a+bi)(a−bi)=a2−b2i2=a2+b2=∣z∣2.\boxed{z\,\bar{z} = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2 = |z|^2.}

This single identity is the engine behind division of complex numbers, and behind the inequality theory.

Properties of conjugate

For all z,w∈Cz, w \in \mathbb{C}:

  1. (zˉ)‾=z\overline{(\bar{z})} = z (involution).
  2. z+w‾=zˉ+wˉ\overline{z + w} = \bar{z} + \bar{w}.
  3. zw‾=zˉ wˉ\overline{zw} = \bar{z}\,\bar{w}.
  4. z/w‾=zˉ/wˉ\overline{z/w} = \bar{z}/\bar{w} (when w≠0w \ne 0).
  5. z+zˉ=2 Re(z)z + \bar{z} = 2\,\text{Re}(z).
  6. z−zˉ=2i Im(z)z - \bar{z} = 2i\,\text{Im}(z).
  7. z=zˉ  ⟺  zz = \bar{z} \iff z is real. z=−zˉ  ⟺  zz = -\bar{z} \iff z is purely imaginary.

Each is proved by direct expansion.

Properties of modulus

For all z,w∈Cz, w \in \mathbb{C}:

  1. ∣z∣≥0|z| \ge 0, with equality iff z=0z = 0.
  2. ∣z∣=∣zˉ∣|z| = |\bar{z}|.
  3. ∣zw∣=∣z∣⋅∣w∣|zw| = |z| \cdot |w| (multiplicative).
  4. ∣zw∣=∣z∣∣w∣\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}.
  5. ∣zn∣=∣z∣n|z^n| = |z|^n.
  6. Triangle inequality: ∣z+w∣≤∣z∣+∣w∣|z + w| \le |z| + |w|.
  7. Reverse triangle: ∣∣z∣−∣w∣∣≤∣z−w∣\big||z| - |w|\big| \le |z - w|.

The multiplicative property is striking: complex multiplication multiplies sizes. The triangle inequality is a direct generalisation of the inequality ∣x+y∣≤∣x∣+∣y∣|x + y| \le |x| + |y| for real numbers.

Proof of ∣zw∣=∣z∣∣w∣|zw| = |z||w|

∣zw∣2=(zw)(zw‾)=(zw)(zˉwˉ)=(zzˉ)(wwˉ)=∣z∣2∣w∣2|zw|^2 = (zw)(\overline{zw}) = (zw)(\bar{z}\bar{w}) = (z\bar{z})(w\bar{w}) = |z|^2 |w|^2.

Taking positive square roots, ∣zw∣=∣z∣∣w∣|zw| = |z||w|. \qed\qed

Computing 1/z1/z

For z≠0z \ne 0: 1z=zˉzzˉ=zˉ∣z∣2=a−bia2+b2.\frac{1}{z} = \frac{\bar{z}}{z \bar{z}} = \frac{\bar{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}.

This is the practical method for dividing complex numbers.

Worked examples

Example 1. Find the modulus and conjugate of z=3−4iz = 3 - 4i.

∣z∣=9+16=5|z| = \sqrt{9 + 16} = 5. zˉ=3+4i\bar{z} = 3 + 4i.

Example 2. Verify ∣zw∣=∣z∣∣w∣|zw| = |z||w| for z=1+iz = 1 + i, w=2−iw = 2 - i.

zw=2−i+2i−i2=3+izw = 2 - i + 2i - i^2 = 3 + i. ∣zw∣=9+1=10|zw| = \sqrt{9 + 1} = \sqrt{10}. ∣z∣=2,∣w∣=5|z| = \sqrt{2}, |w| = \sqrt{5}, product 10\sqrt{10}. Verified.

Example 3. Find 3+2i1−i\dfrac{3 + 2i}{1 - i} using the conjugate technique.

(3+2i)(1+i)(1−i)(1+i)=3+3i+2i+2i21+1=1+5i2=12+52i\dfrac{(3 + 2i)(1 + i)}{(1 - i)(1 + i)} = \dfrac{3 + 3i + 2i + 2i^2}{1 + 1} = \dfrac{1 + 5i}{2} = \dfrac{1}{2} + \dfrac{5}{2} i.

Example 4. Find all zz with z+zˉ=4z + \bar{z} = 4 and zzˉ=13z \bar{z} = 13.

z+zˉ=2Re(z)=4⇒Re(z)=2z + \bar{z} = 2\text{Re}(z) = 4 \Rightarrow \text{Re}(z) = 2. zzˉ=∣z∣2=13z \bar{z} = |z|^2 = 13. If z=2+biz = 2 + bi, then ∣z∣2=4+b2=13⇒b=±3|z|^2 = 4 + b^2 = 13 \Rightarrow b = \pm 3. So z=2+3iz = 2 + 3i or z=2−3iz = 2 - 3i.

Example 5 (harder). Show that if ∣z∣=1|z| = 1, then z+1zz + \dfrac{1}{z} is real.

If ∣z∣=1|z| = 1, then zzˉ=1z\bar{z} = 1, so zˉ=1/z\bar{z} = 1/z. Hence 1z=zˉ\dfrac{1}{z} = \bar{z}, and z+1z=z+zˉ=2Re(z)∈Rz + \dfrac{1}{z} = z + \bar{z} = 2\text{Re}(z) \in \mathbb{R}. \qed\qed

Try it yourself

  1. Find ∣z∣|z| and zˉ\bar{z} for z=−1+2iz = -1 + 2i.
  2. Show ∣3+4i∣=5|3 + 4i| = 5, ∣5−12i∣=13|5 - 12i| = 13.
  3. Find zz such that zzˉ=25z \bar{z} = 25 and z+zˉ=6z + \bar{z} = 6.
  4. Verify the triangle inequality for z=3+4i,w=5−12iz = 3 + 4i, w = 5 - 12i.
  5. Show (z+w)‾=zˉ+wˉ\overline{(z + w)} = \bar{z} + \bar{w} by direct computation.
  6. If z=2+iz = 2 + i, compute 1z\dfrac{1}{z}.
  7. Find all zz with ∣z−1∣=1|z - 1| = 1 and zzˉ=1z \bar{z} = 1.
  8. Prove ∣z∣2+∣w∣2=12(∣z+w∣2+∣z−w∣2)|z|^2 + |w|^2 = \tfrac{1}{2}(|z + w|^2 + |z - w|^2) (parallelogram law).
  9. If ∣z+1∣=∣z−1∣|z + 1| = |z - 1|, prove zz is purely imaginary.
  10. Find all zz with z=zˉ2z = \bar{z}^2.
  11. Show ∣z+w∣2=∣z∣2+∣w∣2+2Re(zwˉ)|z + w|^2 = |z|^2 + |w|^2 + 2\text{Re}(z \bar{w}).
  12. If ∣z1∣=∣z2∣=∣z3∣=1|z_1| = |z_2| = |z_3| = 1 and z1+z2+z3=0z_1 + z_2 + z_3 = 0, prove ∣z1+z2∣=1|z_1 + z_2| = 1.

Pitfalls / Tricks

  • ∣z∣|z| is a real number, never complex. So ∣z∣≥0|z| \ge 0 and ∣z∣2≠z2|z|^2 \ne z^2.
  • z2‾=zˉ2\overline{z^2} = \bar{z}^2, but z2≠∣z∣2z^2 \ne |z|^2 unless zz is real.
  • The triangle inequality is not always tight: equality holds iff z,wz, w lie on the same ray from the origin.
  • Insight. zzˉ=∣z∣2z\bar{z} = |z|^2 is the key identity. Whenever you see zz in a denominator or under a modulus squared, multiply by its conjugate.

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