The complex conjugate and division

What the conjugate is

• The complex conjugate of $z = a + bi$ is written $\overline{z}$ (or sometimes $z^*$) and is formed by flipping the sign of the imaginary part:

$$\overline{a + bi} = a - bi$$

• $\overline{3 + 7i} = 3 - 7i$, $\overline{-2 - i} = -2 + i$, $\overline{5} = 5$.
• The real part is left alone. Only the sign in front of $i$ changes.
• A real number is its own conjugate, because there is no imaginary part to flip.

• Geometrically, taking the conjugate reflects the point in the real axis. That picture explains every property below.

The property that makes conjugates useful

• Multiplying a complex number by its own conjugate always gives a real number:

$$z\overline{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2$$

• $a^2 + b^2$ is a sum of squares, so it is always real and never negative.
• This is the difference of two squares doing the same job it does when rationalising surds — except that here $i^2 = -1$ turns the minus into a plus.
• $(4 + 3i)(4 - 3i) = 16 + 9 = 25$.

Other conjugate properties

• Adding a number to its conjugate leaves twice the real part:

$$z + \overline{z} = 2a = 2\,\text{Re}(z)$$

• Subtracting leaves twice the imaginary part, times $i$:

$$z - \overline{z} = 2bi$$

• The conjugate passes through sums and products unchanged:

  • $$\overline{z + w} = \overline{z} + \overline{w}$$
  • $$\overline{zw} = \overline{z}\;\overline{w}$$

• The consequence that matters for polynomials: if a polynomial has real coefficients and $z$ is a root, then $\overline{z}$ is also a root. Complex roots of real polynomials always come in conjugate pairs.

Dividing complex numbers

• You cannot leave $i$ in a denominator, for the same reason you cannot leave a surd there.
• Multiply the top and bottom by the conjugate of the denominator.

$$\frac{a + bi}{c + di} = \frac{a + bi}{c + di} \times \frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{c^2 + d^2}$$

• The denominator becomes the real number $c^2 + d^2$, and then you simply divide each part.

• The procedure, step by step:

  • Write the conjugate of the denominator — not the numerator.
  • Multiply both top and bottom by it.
  • Expand the numerator, replacing $i^2$ with $-1$.
  • Compute the denominator as $c^2 + d^2$ directly; there is no need to expand it in full.
  • Split into the form $a + bi$ by dividing each part.

• Worked through:

$$\frac{5 + i}{2 - 3i} = \frac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = \frac{10 + 15i + 2i + 3i^2}{4 + 9}$$

$$= \frac{10 + 17i - 3}{13} = \frac{7 + 17i}{13} = \frac{7}{13} + \frac{17}{13}i$$

Simplifying a fraction with $i$ alone underneath

• If the denominator is just $bi$, multiplying by the conjugate still works, but there is a shortcut: multiply top and bottom by $i$.

$$\frac{3}{2i} = \frac{3i}{2i^2} = \frac{3i}{-2} = -\frac{3}{2}i$$

• Equivalently, dividing by $i$ is the same as multiplying by $-i$, because $\dfrac{1}{i} = -i$.

Finding the reciprocal

• The reciprocal $\dfrac{1}{z}$ is a division like any other:

$$\frac{1}{z} = \frac{\overline{z}}{z\overline{z}} = \frac{a - bi}{a^2 + b^2}$$

• So $\dfrac{1}{3 + 4i} = \dfrac{3 - 4i}{25} = \dfrac{3}{25} - \dfrac{4}{25}i$.