Modulus and conjugate

Two new tools

You can now add, subtract and multiply complex numbers. Two more tools make them much easier to handle. One is the conjugate, a partner for each number. The other is the modulus, the size of a number.

The conjugate

For $z = a + ib$, the conjugate is $\bar z = a - ib$. Keep the real part and change the sign of the imaginary part.

Add and subtract a number and its conjugate:

So the real part is $\dfrac{z + \bar z}{2}$. A number is real exactly when $\bar z = z$. It is purely imaginary exactly when $\bar z = -z$.

The modulus

The modulus of $z = a + ib$ is

For example, $|3 + 4i| = \sqrt{9 + 16} = 5$ and $|-5 + 12i| = \sqrt{25 + 144} = 13$. Also $|2 - i| = \sqrt{4 + 1} = \sqrt{5}$.

For a real number $a$, this gives $\sqrt{a^2}$, its usual size. So $|-7| = 7$ as before. The modulus is never negative. It is $0$ only for $z = 0$. Also $|\bar z| = |z|$ and $|-z| = |z|$.

The key product

Multiply a number by its conjugate:

So $z\bar z = |z|^2$. It is always real, and never negative. This one line explains most of what follows.

In the simulation, each number is drawn as the point $(a, b)$. The next lesson explains this picture fully, with the conjugate as a mirror image.

Try $a = 0$. The two points lie on the vertical axis, and $z$ plus its conjugate is $0$.

Rules for the conjugate

For any complex numbers $z_1$ and $z_2$:

• $\overline{\bar z} = z$;
• $\overline{z_1 + z_2} = \bar z_1 + \bar z_2$ and $\overline{z_1 - z_2} = \bar z_1 - \bar z_2$;
• $\overline{z_1 z_2} = \bar z_1 \bar z_2$;
• $\overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{\bar z_1}{\bar z_2}$, when $z_2 \ne 0$.

Check the product rule with $z_1 = 1 + 2i$ and $z_2 = 3 - i$. Here $z_1 z_2 = 3 - i + 6i - 2i^2 = 5 + 5i$, whose conjugate is $5 - 5i$. And $\bar z_1 \bar z_2 = (1 - 2i)(3 + i) = 3 + i - 6i - 2i^2 = 5 - 5i$. They agree.

Rules for the modulus

The modulus of a product is the product of the moduli:

Why? $|z_1 z_2|^2 = (z_1 z_2)\overline{(z_1 z_2)} = z_1 \bar z_1 \, z_2 \bar z_2 = |z_1|^2 |z_2|^2$. Take square roots.

Sums are different. Take $z_1 = 1$ and $z_2 = i$. Then $|z_1 + z_2| = |1 + i| = \sqrt{2}$, but $|z_1| + |z_2| = 2$. In general, $|z_1 + z_2| \le |z_1| + |z_2|$.

Dividing with the conjugate

To divide by $z_2$, multiply the top and the bottom by $\bar z_2$. The bottom becomes $z_2 \bar z_2 = |z_2|^2$, a real number:

In particular, the multiplicative inverse of $z \ne 0$ is $z^{-1} = \dfrac{\bar z}{|z|^2}$. For example, $(3 + 4i)^{-1} = \dfrac{3 - 4i}{25}$.

A worked example

Let us write $\dfrac{3 + 2i}{1 - 4i}$ in the form $a + ib$.

Step 1. Choose the factor. The conjugate of the bottom is $1 + 4i$. Multiply the top and the bottom by it.

Step 2. The bottom. $(1 - 4i)(1 + 4i) = 1^2 + 4^2 = 17$.

Step 3. The top. $(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = -5 + 14i$.

Step 4. Divide. $\dfrac{3 + 2i}{1 - 4i} = -\dfrac{5}{17} + \dfrac{14}{17}i$.

Practise with new numbers

Try it in Python

Python has complex numbers built in. It writes $i$ as j, so $3 + 2i$ is typed 3 + 2j. The program finds a conjugate and a modulus, and divides two ways.

Common mistakes

• Changing the wrong sign. The conjugate of $3 + 2i$ is $3 - 2i$, not $-3 + 2i$. Only the imaginary part changes sign.
• Putting $i$ into the modulus. $|3 + 2i| = \sqrt{9 + 4}$. Do not write $\sqrt{9 + 4i^2}$, which is $\sqrt{5}$.
• Adding moduli for a sum. $|z_1 + z_2|$ is usually less than $|z_1| + |z_2|$.
• Multiplying by the wrong number. To divide by $1 - 4i$, multiply the top and the bottom by $1 + 4i$.

Remember these

• $\bar z = a - ib$ and $|z| = \sqrt{a^2 + b^2}$ for $z = a + ib$.
• $z\bar z = |z|^2$, a real number that is never negative.
• Conjugates pass through sums, differences, products and quotients.
• $|z_1 z_2| = |z_1||z_2|$, but $|z_1 + z_2|$ is at most $|z_1| + |z_2|$.
• $\dfrac{z_1}{z_2} = \dfrac{z_1 \bar z_2}{|z_2|^2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write the conjugate and the modulus of $-6 - 8i$.

2. Find $z\bar z$ for $z = 5 - 2i$ without multiplying out the brackets.

3. The moduli of two numbers are $3$ and $4$. Which can you find from these alone: the modulus of their product, or of their sum?

4. Write $\dfrac{1}{2 + i}$ in the form $a + ib$.

5. Show that $z + \bar z$ is always a real number.