Paul's Online Math Notes
Complex Number Primer / Conjugate and Modulus  
Complex Number Primer
Arithmetic Previous Chapter   Next Chapter Polar and Exponential Forms

 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  the complex conjugate is denoted by  and is defined to be,

 

(1)

 

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

 

Here are some basic facts about conjugates.

 

(2)

 

(3)

 

(4)

 

(5)

 

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.

(a)  for  

(b)  for  and  

(c)  for  and  

Solution

There really isn’t much to do with these other than to so the work so,

(a)  

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

 

(b)  

 

(c)  

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  and then perform each of the following operations.

 

 

Now, recalling that  and  we see that we have,

(6)

Modulus

The other operation we want to take a look at in this section is the modulus of a complex number.  Given a complex number  the modulus is denoted by  and is defined by

(7)

 

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.  ) then,

 

 

where the  on the z is the modulus of the complex number and the  on the a is the absolute value of a real number (recall that in general for any real number a we have  ).  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 numbers and its real and imaginary parts.  To see this let’s square both sides of (7) and use the fact that  and .  Doing this we arrive at

 

 

Since all three of these terms are positive we can drop the Im z part on the left which gives the following inequality,

 

 

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

 

 

where the  on the z is the modulus of the complex number and the  on the Re z are absolute value bars.  Finally, for any real number a we also know that  (absolute value…) and so we get,

(8)

 

We can use a similar argument to arrive at,

(9)

 

There is a very nice relationship between the modulus of a complex number and it’s conjugate.  Let’s start with a complex number  and take a look at the following product.

 

 

From this product we can see that

(10)

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,

(11)

and

(12)

 

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 (10) we can simplify the notation a little.  Doing all this gives the following formula for division,

 

 

 

Here’s a quick example illustrating this,

 

Example 2  Evaluate .

Solution

In this case we have  and .  Then computing the various parts of the formula gives,

                                       

The quotient is then,

                          

 

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

 

(13)

 

(14)

 

(15)

 

Property (13) should make some sense to you.  If the modulus is zero then , but the only way this can be zero is if both a and b are zero.

 

To verify (14) consider the following,

 

 

 

 

So, from this we can see that

 

 

 

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

 

 

 

Property (15) can be verified using a similar argument.

 

Triangle Inequality and Variants

Properties (14) and (15) 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,

(16)

 

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 (10) and (3) to rewrite it a little.

 

 

Now multiply out the right side to get,

(17)

Next notice that,

 

 

and so using (6), (8) and (11) we can write middle two terms of the right side of (17) as

 

 

 

Also use (10) on the first and fourth term in (17) to write them as,

 

 

 

With the rewrite on the middle two terms we can now write (17) as

 

 

So, putting all this together gives,

 

 

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

 

 

 

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

 

Let’s first start by assuming that .  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  and do some work on it.

 

 

 

Now, rewrite things a little and we get,

(18)

 

If we now assume that  we can go through a similar process as above except this time switch  and  and we get,

(19)

 

Now, recalling the definition of absolute value we can combine (18) and (19) into the following variation of the triangle inequality.

(20)

 

Also, if we replace  with  in (16) and (20) we arrive at two more variations of the triangle inequality.

(21)

 

(22)

 

On occasion you’ll see (22) called the reverse triangle inequality.

 

Arithmetic Previous Chapter   Next Chapter Polar and Exponential Forms

Complex Number Primer / Conjugate and Modulus   

© 2003 - 2016 Paul Dawkins