Complex analysis is calculus with complex numbers, and two tools from earlier courses do most of the bookkeeping. In the complex plane you met the modulus $|a + bi| = \sqrt{a^2 + b^2}$, the distance from $0$, and the conjugate $a - bi$, the mirror image across the real axis. In complex division you multiplied by the conjugate to clear $i$ from a denominator. This lesson makes both tools exact and quick, because every later topic leans on them.
Key idea
For $z = x + iy$ (with $x, y$ real), the conjugate is $\overline{z} = x - iy$ and the modulus is $|z| = \sqrt{x^2 + y^2}$. Then
$$z\overline{z} = |z|^2, \qquad |zw| = |z|\,|w|, \qquad \left|\frac{z}{w}\right| = \frac{|z|}{|w|}, \qquad |z + w| \le |z| + |w|.$$
Also $\overline{z + w} = \overline{z} + \overline{w}$, $\overline{zw} = \overline{z}\,\overline{w}$, $\operatorname{Re} z = \frac{z + \overline{z}}{2}$ and $\operatorname{Im} z = \frac{z - \overline{z}}{2i}$.
Why it works
Multiply out: $z\overline{z} = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 + y^2$, a real number that is never negative. That one fact gives the rest. For products, $|zw|^2 = zw\,\overline{zw} = (z\overline{z})(w\overline{w}) = |z|^2|w|^2$. For the triangle inequality,
$$|z + w|^2 = (z + w)(\overline{z} + \overline{w}) = |z|^2 + 2\operatorname{Re}(z\overline{w}) + |w|^2 \le |z|^2 + 2|z|\,|w| + |w|^2 = (|z| + |w|)^2,$$
because the real part of a number is at most its modulus. Geometrically: one side of a triangle is never longer than the other two together.
Worked examples
Worked example: Modulus, conjugate and square
Let $z = 3 - 4i$. Find $\overline{z}$, $|z|$, $|z|^2$ and $z^2$.
- Flip the sign of the imaginary part: $\overline{z} = 3 + 4i$.
- $|z|^2 = z\overline{z} = 3^2 + 4^2 = 25$, so $|z| = 5$.
- $z^2 = (3 - 4i)^2 = 9 - 24i + 16i^2 = -7 - 24i$.
- Compare: $z^2$ is not real, while $|z|^2 = 25$ is. (Their sizes agree: $|z^2| = \sqrt{49 + 576} = 25 = |z|^2$.)
Answer: $\overline{z} = 3 + 4i$, $|z| = 5$, $|z|^2 = 25$, $z^2 = -7 - 24i$
Worked example: A reciprocal
Write $\dfrac{1}{2 + i}$ as $a + bi$.
- Use $\dfrac{1}{z} = \dfrac{\overline{z}}{z\overline{z}} = \dfrac{\overline{z}}{|z|^2}$.
- $\overline{z} = 2 - i$ and $|z|^2 = 4 + 1 = 5$.
- $\dfrac{1}{2 + i} = \dfrac{2 - i}{5} = \dfrac{2}{5} - \dfrac{1}{5}i$. Check: $(2 + i)\left(\frac{2}{5} - \frac{1}{5}i\right) = \frac{4}{5} + \frac{1}{5} = 1$.
Answer: $\frac{2}{5} - \frac{1}{5}i$
Worked example: A modulus without multiplying
Find $|(1 + 2i)(3 - 4i)|$.
- $|zw| = |z|\,|w|$, so find the two moduli separately.
- $|1 + 2i| = \sqrt{5}$ and $|3 - 4i| = 5$.
- The product has modulus $5\sqrt{5}$. (Multiplying out gives $11 + 2i$, and $\sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}$: the same.)
Answer: $5\sqrt{5}$
Common mistake
Treating $z^2$ and $|z|^2$ as the same. For $z = 1 + i$, $z^2 = 1 + 2i + i^2 = 2i$, but $|z|^2 = z\overline{z} = 1 + 1 = 2$. The square of $z$ doubles its angle; the square of its modulus is a length squared. Quick check: $|z|^2$ is always a real number $\ge 0$. If your answer for $|z|^2$ still contains $i$, you multiplied $z$ by itself instead of by $\overline{z}$.
The triangle inequality also runs backwards: $|z + w| \ge \big||z| - |w|\big|$. So if $|z| = 3$ and $|w| = 5$, then $|z + w|$ can be anything from $2$ to $8$, and both ends happen. These two bounds are the main tool for estimating integrals later in the course.
Where these identities get used
- Dividing: multiply the top and bottom by $\overline{w}$, so the bottom becomes the real number $|w|^2$.
- Solving for $z$: an equation that mixes $z$ and $\overline{z}$ is two real equations; write $z = x + iy$ and match real and imaginary parts.
- Estimating: to bound $\left|\frac{1}{z^2 + 4}\right|$ on the circle $|z| = 3$, use $|z^2 + 4| \ge |z|^2 - 4 = 5$, so the fraction is at most $\frac{1}{5}$.