# What is a conjugate?

• Apr 26th 2009, 11:13 PM
TheBerkeleyBoss
What is a conjugate?
Throughout the different math classes I have taken, I've noticed that a few mathematical concepts were slightly "sidelined". Which brings me to my question....I've always heard teachers mentioning conjugates. I was wondering if anyone could shine any light on what a conjugate is and how it could help me solve algebraic equations. (and complex conjugates if you have time!)

Thanks!

-Andy-
• Apr 27th 2009, 01:52 AM
Prove It
Quote:

Originally Posted by TheBerkeleyBoss
Throughout the different math classes I have taken, I've noticed that a few mathematical concepts were slightly "sidelined". Which brings me to my question....I've always heard teachers mentioning conjugates. I was wondering if anyone could shine any light on what a conjugate is and how it could help me solve algebraic equations. (and complex conjugates if you have time!)

Thanks!

-Andy-

If you have a binomial, in other words, a sum of two terms

$a + b$

It's conjugate is simply another binomial with the same numbers, but the sign separating them changes.

So the conjugate of $a + b$ is $a - b$.

The conjugate is an extremely powerful tool because usually when you expand two binomials you get

$(a + b)(c + d) = ab + ad + bc + bd$

but when you multiply conjugates, the middle terms cancel.

$(a + b)(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2$.

This is known as the difference of two squares.

This property comes in handy if you are dealing with surds, in particular, when you have surds in the denominator and want to rewrite it in a nicer form.

E.g. $\frac{1}{3 + \sqrt{2}} = \frac{1}{3 + \sqrt{2}} \times \color{red}{1}$

$= \frac{1}{3 + \sqrt{2}} \times \color{red}{\frac{3 - \sqrt{2}}{3 - \sqrt{2}}}$

$= \frac{3 - \sqrt{2}}{3 - (\sqrt{2})^2}$

$= \frac{3 - \sqrt{2}}{3 - 2}$

$= \frac{3 - \sqrt{2}}{1}$

$= 3 - \sqrt{2}$, which is a MUCH nicer form.

A similar process is used when dividing complex numbers.

For a complex number $a + ib$, it's complex conjugate is $a - ib$.

Suppose you had $\frac{2 + 3i}{1 + i}$.

To work out this division, you multiply top and bottom by its complex conjugate, similar to how you rationalise a denominator.

$\frac{2 + 3i}{1 + i} = \frac{2 + 3i}{1 + i} \times \color{red}{1}$

$= \frac{2 + 3i}{1 + i} \times \color{red}{\frac{1 - i}{1 - i}}$

$= \frac{(2 + 3i)(1 - i)}{1^2 - i^2}$

$= \frac{2 - 2i + 3i - 3i^2}{1 + 1}$

$= \frac{5 + i}{2}$

$= \frac{5}{2} + \frac{1}{2}i$.

The other nice property of complex conjugates is the fact that when you multiply a complex number by its conjugate, you get the square of the length of the complex number itself. This makes it very easy to calculate the length of a complex number.

Let $z = a + ib$. It's conjugate is therefore $z^{-} = a - ib$

$zz^{-} = (a + ib)(a - ib)$

$= a^2 - i^2b^2$

$= a^2 + b^2$.

The length of $z$ is $|z| = \sqrt{a^2 + b^2}$ (by Pythagoras).

So $|z|^2 = a^2 + b^2 = zz^{-}$.

Hope that helped.
• Feb 14th 2010, 09:09 PM
JohnD1079
Quote:

Originally Posted by Prove It

$= \frac{3 - \sqrt{2}}{3 - (\sqrt{2})^2}$

$= \frac{3 - \sqrt{2}}{3 - 2}$

$= \frac{(2 + 3i)(1 - i)}{1^2 - i^2}$

$= \frac{2 - 2i + 3i - 3i^2}{1 + 1}$

Can someone explain what happened here?
For the 1st one, shouldnt the denominator be 9-4=5?
• Feb 14th 2010, 11:14 PM
Prove It
Quote:

Originally Posted by JohnD1079
Can someone explain what happened here?
For the 1st one, shouldnt the denominator be 9-4=5?

Yes it is, my mistake...
• Feb 15th 2010, 03:12 AM
mathemagister
Quote:

Originally Posted by Prove It
Yes it is, my mistake...

I'm confused; shouldn't it be 9-2=7 not 9-4=5??
• Feb 15th 2010, 03:15 AM
Prove It
Oh for the Love of...

Yes, the denominator becomes

$(3 + \sqrt{2})(3 - \sqrt{2})$

$= 3^2 - (\sqrt{2})^2$

$= 9 - 2$

$= 7$.

Why was a thread that's nearly a year old bumped anyway?
• Feb 15th 2010, 05:20 AM
HallsofIvy
Because they always come back to haunt you!(Rofl)
• Jul 27th 2010, 08:48 AM
bwpbruce
Which is simpler and why?
(Rofl)