Paul's Online Notes
Home / Complex Number Primer / Conjugate and Modulus
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

## Conjugate and Modulus

In the previous section we looked at algebraic operations on complex numbers.There are a couple of other operations that we should take a look at since they tend to show up on occasion.We’ll also take a look at quite a few nice facts about these operations.

#### Complex Conjugate

The first one we’ll look at is the complex conjugate, (or just the conjugate).Given the complex number $$z = a + bi$$ the complex conjugate is denoted by $$\overline z$$ and is defined to be,

\begin{equation}\overline z = a - bi\end{equation}

In other words, we just switch the sign on the imaginary part of the number.

Here are some basic facts about conjugates.

\begin{align} \overline{\overline{z}} & = z \\ \overline {{z_1} \pm {z_2}} & = {\overline z_1} \pm {\overline z_2} \label{eq:conjsum} \\ \overline {{z_1}{z_2}} &= {\overline z_1}\,{\overline z_2} \label{eq:conjprod} \\ \overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} & = \frac{{{{\overline z}_1}}}{{{{\overline z}_2}}} \end{align}

The first one just says that if we conjugate twice we get back to what we started with originally and hopefully this makes some sense.The remaining three just say we can break up sum, differences, products and quotients into the individual pieces and then conjugate.

So, just so we can say that we worked a number example or two let’s do a couple of examples illustrating the above facts.

Example 1 Compute each of the following.
1. $$\overline{\overline{z}}$$ for $$z = 3 - 15i$$
2. $$\overline {{z_1} - {z_2}}$$ for $${z_1} = 5 + i$$ and $${z_2} =- 8 + 3i$$
3. $${\overline{z_1}} - {\overline{z_2}}$$ for $${z_1} = 5 + i$$ and $${z_2} =- 8 + 3i$$
Show Solution
There really isn’t much to do with these other than to so the work so,

a $$\overline{z} = 3 + 15i \hspace{0.5in} \Rightarrow \hspace{0.5in} \overline{\overline{z}} = \overline{3 + 15i}= 3 - 15i = z$$

Sure enough we can see that after conjugating twice we get back to our original number.

b $${z_1} - {z_2} = 13 - 2i \hspace{0.5in} \Rightarrow \hspace{0.5in} \overline {{z_1} - {z_2}}= \overline {13 - 2i}= 13 + 2i$$

c $${\overline z_1} - {\overline z_2} = \overline {5 + i}- \left( {\overline { - 8 + 3i} } \right) = 5 - i - \left( { - 8 - 3i} \right) = 13 + 2i$$

We can see that results from (b) and (c) are the same as the fact implied they would be.

There is another nice fact that uses conjugates that we should probably take a look at.However, instead of just giving the fact away let’s derive it.We’ll start with a complex number $$z = a + bi$$ and then perform each of the following operations.

$\begin{array}{rlcrl}z + \overline{z} & = a + bi + \left( {a - bi} \right) & \hspace{0.5in} & z - \overline{z} & = a + bi - \left( {a - bi} \right)\\ & = 2a & \hspace{0.5in} & & = 2bi\end{array}$

Now, recalling that $${\mathop{\rm Re}\nolimits}\, z = a$$ and $${\mathop{\rm Im}\nolimits}\, z = b$$ we see that we have,

\begin{equation}{\mathop{\rm Re}\nolimits} z = \frac{{z + \overline z}}{2} \hspace{0.75in} {\mathop{\rm Im}\nolimits} z = \frac{{z - \overline z}}{{2i}} \label{eq:ReImDefn} \end{equation}

#### Modulus

TThe other operation we want to take a look at in this section is the modulus of a complex number.Given a complex number $$z = a + bi$$ the modulus is denoted by $$\left| z \right|$$ and is defined by

\begin{equation}\left| z \right| = \sqrt {{a^2} + {b^2}} \label{eq:ModDefn} \end{equation}

Notice that the modulus of a complex number is always a real number and in fact it will never be negative since square roots always return a positive number or zero depending on what is under the radical.

Notice that if $$z$$ is a real number (i.e. $$z = a + 0i$$) then,

$\left| z \right| = \sqrt {{a^2}}= \left| a \right|$

where the $$\left| {\,\, \cdot \,} \right|$$ on the $$z$$ is the modulus of the complex number and the $$\left| {\,\, \cdot \,} \right|$$ on the $$a$$ is the absolute value of a real number (recall that in general for any real number $$a$$ we have $$\sqrt {{a^2}}= \left| a \right|$$).So, from this we can see that for real numbers the modulus and absolute value are essentially the same thing.

We can get a nice fact about the relationship between the modulus of a complex number and its real and imaginary parts.To see this let’s square both sides of $$\eqref{eq:ModDefn}$$ and use the fact that $${\mathop{\rm Re}\nolimits}\, z = a$$ and $${\mathop{\rm Im}\nolimits}\, z = b$$.Doing this we arrive at

${\left| z \right|^2} = {a^2} + {b^2} = {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2} + {\left( {{\mathop{\rm Im}\nolimits}\, z} \right)^2}$

Since all three of these terms are positive we can drop the $${\mathop{\rm Im}\nolimits}\, z$$ part on the left which gives the following inequality,

${\left| z \right|^2} = {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2} + {\left( {{\mathop{\rm Im}\nolimits}\, z} \right)^2} \ge {\left( {{\mathop{\rm Re}\nolimits}\, z} \right)^2}$

If we then square root both sides of this we get,

$\left| z \right| \ge \left| {{\mathop{\rm Re}\nolimits}\, z} \right|$

where the $$\left| {\,\, \cdot \,} \right|$$ on the $$z$$ is the modulus of the complex number and the $$\left| {\,\, \cdot \,} \right|$$ on the $${\mathop{\rm Re}\nolimits}\, z$$ are absolute value bars. Finally, for any real number $$a$$ we also know that $$a \le \left| a \right|$$ (absolute value…) and so we get,

\begin{equation}\left| z \right| \ge \left| {{\mathop{\rm Re}\nolimits}\, z} \right| \ge {\mathop{\rm Re}\nolimits}\, z \label{eq:zRez} \end{equation}

We can use a similar argument to arrive at,

\begin{equation}\left| z \right| \ge \left| {{\mathop{\rm Im}\nolimits}\, z} \right| \ge {\mathop{\rm Im}\nolimits}\, z \label{eq:zImz} \end{equation}

There is a very nice relationship between the modulus of a complex number and its conjugate.Let’s start with a complex number $$z = a + bi$$ and take a look at the following product.

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

From this product we can see that

\begin{equation}z\,\overline z = {\left| z \right|^2}\label{eq:zConjz} \end{equation}

This is a nice and convenient fact on occasion.

Notice as well that in computing the modulus the sign on the real and imaginary part of the complex number won’t affect the value of the modulus and so we can also see that,

\begin{equation}\left| z \right| = \left| {\overline z} \right|\label{eq:MzMzbar}\end{equation}

and

\begin{equation}\left| { - z} \right| = \left| z \right|\end{equation}

We can also now formalize the process for division from the previous section now that we have the modulus and conjugate notations.In order to get the $$i$$ out of the denominator of the quotient we really multiplied the numerator and denominator by the conjugate of the denominator.Then using $$\eqref{eq:zConjz}$$ we can simplify the notation a little.Doing all this gives the following formula for division,

$\frac{{{z_1}}}{{{z_2}}} = \frac{{{z_1}\,{{\overline z}_2}}}{{{z_2}\,{{\overline z}_2}}} = \frac{{{z_1}\,{{\overline z}_2}}}{{{{\left| {{z_2}} \right|}^2}}}$

Here’s a quick example illustrating this,

Example 2 Evaluate $$\frac{{6 + 3i}}{{10 + 8i}}$$.
Show Solution
In this case we have $${z_1} = 6 + 3i$$ and $${z_2} = 10 + 8i$$.Then computing the various parts of the formula gives, ${\overline z_2} = 10 - 8i \hspace{0.75in} {\left| {{z_2}} \right|^2} = {10^2} + {8^2} = 164$

The quotient is then,

$\frac{{6 + 3i}}{{10 + 8i}} = \frac{{\left( {6 + 3i} \right)\left( {10 - 8i} \right)}}{{164}} = \frac{{60 - 48i + 30i - 24{i^2}}}{{164}} = \frac{{21}}{{41}} - \frac{9}{{82}}i$

Here are some more nice facts about the modulus of a complex number.

\begin{align}{\rm{If }}\left| z \right| & = 0\,\,\,{\rm{then}}\, z = 0 \label{eq:Mzero} \\ \left| {{z_1}\,{z_2}} \right| & = \left| {{z_1}} \right|\,\left| {{z_2}} \right| \label{eq:MProd} \\ \left| {\frac{{{z_1}}}{{{z_2}}}} \right| & = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} \label{eq:MQuot} \end{align}

Property $$\eqref{eq:Mzero}$$ should make some sense to you.If the modulus is zero then $${a^2} + {b^2} = 0$$, but the only way this can be zero is if both $$a$$ and $$b$$ are zero.

To verify $$\eqref{eq:MProd}$$ consider the following,

\begin{align*}{\left| {{z_1}\,{z_2}} \right|^2} & = \left( {{z_1}\,{z_2}} \right)\overline {\left( {{z_1}\,{z_2}} \right)} & \hspace{0.5in} & {\text{using property }} \eqref{eq:zConjz} \\ & = \left( {{z_1}\,{z_2}} \right)\left( {{{\overline z}_1}\,{{\overline z}_2}} \right) & \hspace{0.5in} & {\text{using property }} \eqref{eq:conjprod} \\ & = {z_1}{{\overline z}_1}\,{z_2}{{\overline z}_2} & \hspace{0.5in} & {\text{rearranging terms}}\\ & = {\left| {{z_1}} \right|^2}\,{\left| {{z_2}} \right|^2} & \hspace{0.5in} & {\text{using property \eqref{eq:zConjz} again (twice)}}\end{align*}

So, from this we can see that

${\left| {{z_1}\,{z_2}} \right|^2} = {\left| {{z_1}} \right|^2}\,{\left| {{z_2}} \right|^2}$

Finally, recall that we know that the modulus is always positive so take the square root of both sides to arrive at

$\left| {{z_1}\,{z_2}} \right| = \left| {{z_1}} \right|\,\,\left| {{z_2}} \right|$

Property $$\eqref{eq:MQuot}$$ can be verified using a similar argument.

Triangle Inequality and Variants

Properties $$\eqref{eq:MProd}$$ and $$\eqref{eq:MQuot}$$ relate the modulus of a product/quotient of two complex numbers to the product/quotient of the modulus of the individual numbers.We now need to take a look at a similar relationship for sums of complex numbers.This relationship is called the triangle inequality and is,

\begin{equation}\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right| \label{eq:triangle} \end{equation}

We’ll also be able to use this to get a relationship for the difference of complex numbers.

The triangle inequality is actually fairly simple to prove so let’s do that. We'll start with the left side squared and use $$\eqref{eq:zConjz}$$ and $$\eqref{eq:conjsum}$$ to rewrite it a little.

${\left| {{z_1} + {z_2}} \right|^2} = \left( {{z_1} + {z_2}} \right)\left( {\overline {{z_1} + {z_2}} } \right) = \left( {{z_1} + {z_2}} \right)\left( {\overline {{z_1}}+ \overline {{z_2}} } \right)$

Now multiply out the right side to get,

\begin{equation}{\left| {{z_1} + {z_2}} \right|^2} = {z_1}\,{\overline z_1} + {z_1}\,{\overline z_2} + {z_2}\,{\overline z_1} + {z_2}\,{\overline z_2}\label{eq:tripfone} \end{equation}

Next notice that,

$\overline {{z_2}{{\overline z}_1}}= {\overline z_2}{\overline{\overline{z_1}}} = {\overline z_2}{z_1}$

and so using $$\eqref{eq:ReImDefn}$$, $$\eqref{eq:zRez}$$ and $$\eqref{eq:MzMzbar}$$ we can write middle two terms of the right side of $$\eqref{eq:tripfone}$$ as

${z_1}\,{\overline z_2} + {z_2}\,{\overline z_1} = {z_1}\,{\overline z_2} + \overline {{z_1}\,{{\overline z}_2}}= 2{\mathop{\rm Re}\nolimits} \left( {{z_1}\,{{\overline z}_2}} \right) \le 2\left| {{z_1}\,{{\overline z}_2}} \right| = 2\left| {{z_1}} \right|\,\left| {{{\overline z}_2}} \right| = 2\left| {{z_1}} \right|\,\left| {{z_2}} \right|$

Also use $$\eqref{eq:zConjz}$$ on the first and fourth term in $$\eqref{eq:tripfone}$$ to write them as,

${z_1}\,{\overline z_1} = {\left| {{z_1}} \right|^2} \hspace{0.75in} {z_2}\,{\overline z_2} = {\left| {{z_2}} \right|^2}$

With the rewrite on the middle two terms we can now write $$\eqref{eq:tripfone}$$ as

\begin{align*}{\left| {{z_1} + {z_2}} \right|^2} & = {z_1}\,{{\overline z}_1} + {z_1}\,{{\overline z}_2} + {z_2}\,{{\overline z}_1} + {z_2}\,{{\overline z}_2}\\ & = {\left| {{z_1}} \right|^2} + {z_1}\,{{\overline z}_2} + {z_2}\,{{\overline z}_1} + {\left| {{z_2}} \right|^2}\\ & \le {\left| {{z_1}} \right|^2} + 2\left| {{z_1}} \right|\left| {{z_2}} \right| + {\left| {{z_2}} \right|^2}\\ & = {\left( {\left| {{z_1}} \right| + \left| {{z_2}} \right|} \right)^2}\end{align*}

So, putting all this together gives,

${\left| {{z_1} + {z_2}} \right|^2} \le {\left( {\left| {{z_1}} \right| + \left| {{z_2}} \right|} \right)^2}$

Now, recalling that the modulus is always positive we can square root both sides and we’ll arrive at the triangle inequality.

$\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$

There are several variations of the triangle inequality that can all be easily derived.

Let’s first start by assuming that $$\left| {{z_1}} \right| \ge \left| {{z_2}} \right|$$.This is not required for the derivation, but will help to get a more general version of what we’re going to derive here.So, let’s start with $$\left| {{z_1}} \right|$$ and do some work on it.

\begin{align*}\left| {{z_1}} \right| & = \left| {{z_1} + {z_2} - {z_2}} \right| & \\ & \le \left| {{z_1} + {z_2}} \right| + \left| { - {z_2}} \right| & \hspace{0.25in} {\text{Using triangle inequality}}\\ & = \left| {{z_1} + {z_2}} \right| + \left| {{z_2}} \right| & \end{align*}

Now, rewrite things a little and we get,

\begin{equation}\left| {{z_1} + {z_2}} \right| \ge \left| {{z_1}} \right| - \left| {{z_2}} \right| \ge 0\label{eq:revtrione} \end{equation}

If we now assume that $$\left| {{z_1}} \right| \le \left| {{z_2}} \right|$$ we can go through a similar process as above except this time switch $${z_1}$$ and $${z_2}$$ and we get,

\begin{equation}\left| {{z_1} + {z_2}} \right| \ge \left| {{z_2}} \right| - \left| {{z_1}} \right| =- \left( {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right) \ge 0\label{eq:revtritwo}\end{equation}

Now, recalling the definition of absolute value we can combine $$\eqref{eq:revtrione}$$ and $$\eqref{eq:revtritwo}$$ into the following variation of the triangle inequality.

\begin{equation}\left| {{z_1} + {z_2}} \right| \ge \left| {{\kern 1pt} \left| {{z_1}} \right| - \left| {{z_2}} \right|{\kern 1pt} } \right|\label{eq:revtrithree}\end{equation}

Also, if we replace $${z_2}$$ with $$- {z_2}$$ in $$\eqref{eq:triangle}$$ and $$\eqref{eq:revtrithree}$$ we arrive at two more variations of the triangle inequality.

\begin{equation}\left| {{z_1} - {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|\end{equation} \begin{equation}\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right| \label{eq:revtri} \end{equation}

On occasion you’ll see $$\eqref{eq:revtri}$$ called the reverse triangle inequality.