The Complex Conjugate
Take any complex number z = a + bi and flip the sign of its
imaginary part. The result is the complex conjugate, written
\bar{z}:
\overline{a + bi} = a - bi.
Geometrically, conjugation is a reflection across the real axis: the point
(a, b) mirrors to (a, -b). This one
tiny operation is the workhorse of complex arithmetic — it is what lets us
divide complex numbers, and what turns z\bar{z}
into a plain real number.
The properties that make it useful
Every rule below follows from the definition by ordinary algebra. The two most important —
conjugate of a sum and conjugate of a product — say that conjugation passes
straight through addition and multiplication, which is exactly why it behaves so well.
For complex numbers z = a + bi and w:
-
conjugate of a sum is the sum of conjugates:
\overline{z + w} = \bar{z} + \bar{w};
-
conjugate of a product is the product of conjugates:
\overline{z\,w} = \bar{z}\,\bar{w};
-
z\bar{z} = |z|^2 = a^2 + b^2 — always real and
\ge 0;
-
z + \bar{z} = 2a = 2\operatorname{Re}(z);
-
z - \bar{z} = 2bi = 2i\operatorname{Im}(z);
-
conjugating twice returns the original: \overline{\bar{z}} = z;
-
z is real \iff z = \bar{z}.
To see the key one, multiply z by \bar{z}
using the
multiplication rule with c = a and
d = -b: the cross terms cancel, i^2 = -1
flips the last sign, and
z\bar{z} = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2.
Dividing complex numbers
Division is the reason the conjugate earns its keep. A quotient
\dfrac{a + bi}{c + di} has an i stuck in
the denominator, and we cannot leave it there. The trick: multiply top and bottom by the
conjugate of the denominator, which clears the i below.
Step 1 — multiply by the conjugate of the denominator. Multiplying by
\dfrac{c - di}{c - di} = 1 changes nothing but the form:
\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)}.
Step 2 — the denominator becomes real. By the
z\bar{z} = c^2 + d^2 rule, the bottom is now a plain real number:
(c + di)(c - di) = c^2 + d^2.
Step 3 — expand the numerator and split into parts. Multiply out the top and
collect real and imaginary terms:
\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}.
That is a clean complex number again — real part over c^2 + d^2,
imaginary part over c^2 + d^2.
A worked example
Compute \dfrac{3 + 2i}{1 + 4i}.
Step 1 — multiply top and bottom by 1 - 4i (the
conjugate of the denominator):
\frac{3 + 2i}{1 + 4i} = \frac{(3 + 2i)(1 - 4i)}{(1 + 4i)(1 - 4i)}.
Step 2 — the denominator. Here
c^2 + d^2 = 1^2 + 4^2 = 17:
(1 + 4i)(1 - 4i) = 1 + 16 = 17.
Step 3 — the numerator. Expand and use i^2 = -1:
(3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i.
Step 4 — assemble. Put numerator over denominator:
\frac{3 + 2i}{1 + 4i} = \frac{11 - 10i}{17} = \frac{11}{17} - \frac{10}{17}i.
The most useful special case is dividing 1 by
z. Multiply top and bottom by \bar{z}:
\frac{1}{z} = \frac{\bar{z}}{z\bar{z}} = \frac{\bar{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}.
So the reciprocal of z is its conjugate scaled down by
|z|^2. For example
\dfrac{1}{3 + 4i} = \dfrac{3 - 4i}{25} = \dfrac{3}{25} - \dfrac{4}{25}i,
since |3 + 4i|^2 = 25.
See the reflection
Slide a and b to move
z = a + bi. Its conjugate \bar{z} = a - bi
is the mirror image across the real axis — same real part, opposite imaginary part. The
readout shows z\bar{z} = a^2 + b^2 recomputed live, always a
non-negative real number.