What is this

**Conjugate**?

Let's me start with an example.

A maths equation can be expressed as Z = a + b.

The

**conjugate**to this Z is "a - b". Let us express this conjugate as conj(Z).

Look closely, we can see that the conjugate is just negating on of the term in Z, that is, the "b" term.

**Note**:

- We can also negate the "a" term but keeping the "b" term untouched.
- The rule is that only one term is reversed in sign.
- The term to be negated is selected on a case-by-case condition to suit the application and objective.

Another example:

S = A + C + F

Conj(S) = (A + C ) - F

Similarly it can be, Conj(S) = A - (C + F)

Again to emphasize, to which conj(S) to use is entirely on its objective of usage.

More Example:

Y= E - G

Conj(Y) = E + G

It can be also Conj(Y) = -E - G.

The message is that we can select one term (or a group of terms) to be reversed in sign.

This is conjugating.

**Its Application**

After knowing what is "Conjugate", let us look at how to apply it.

Let Z = a + b.

If we multiply Z with its conjugate, that is, (Z) x Conj(Z), we will get (a + b) (a - b).

And (a + b)(a - b) = a

^{2}- b

^{2}, which is a special product in algebra.

The usefulness of this application is more obvious in Complex Number computation.

Let H = a + ib

The conjugate is conj(H) = a - ib.

Their product is (H) x conj(H) = (a + ib) (a - ib) ==> a

^{2}- (ib)

^{2}

Rewriting the above, we get a

^{2}- i

^{2}b

^{2}.

Note from complex number understanding, i

^{2}= -1.

Therefore a

^{2}- i

^{2}b

^{2 }= a

^{2}- (-1) b

^{2 }= a

^{2}+ b

^{2 }.

What happened to the result of the product?

We can see that the application of conjugate multiplication results in converting a Complex Number (a + ib) into a

**real**number (without any "i") !

This knowledge when applied properly is very powerful in the

**Division operation**of Complex Number.

As a rule on Conjugate Multiplication of Complex Number:

**Z x Conj(Z) = a**

^{2}+ b^{2}

^{}However, care has to be taken when applying Conjugate.

**Common Mistake in Conjugate**

To perform a conjugate, we only negate one term (or one group of terms).

__Mistake 1__: Z = a + ib ==> Conj(Z) = -a - ib (wrongly done!) ==> Both terms negated !

Complex Number Conjugate Multiplication results in

**a**.

^{2}+ b^{2}__Mistake 2__: Z x Conj(Z) = a

^{2}

**-**b

^{2}==> wrong sign for the second term !

This was confused with the normal algebra special product (a + b)(a - b) = a

^{2}- b

^{2}.

Why such mistake?

It is due to the "i" symbol in Complex Number that actually causes the sign to be negated.

(i

^{2}= -1). The only solution to this mistake is a full awareness of the "i" symbol and understanding what you are doing.

Just a littel "i" can cause so much trouble, right?

On the other hand, what so difficult about handling this "i"? It is just a little "i", that's all.

You have a choice of which thinking and approach you want.

A wise man however will go for a positive one.

:-)

## 3 comments:

Wow that has made it a LOT clearer!! Why can't teachers teach like this =S... They make everything so confusing! D=

Great help!

Nice to knw that this post helps and reduces the confusion for my readers.

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