The Mathematical Idea of Conjugate

The simple factoring

A^2 – B^2=(A + B)(A – B)

has lots of applications. The fact that a "difference of squares" factors into a sum times a difference –note the use of three different basic operations! – means that we can often manipulate square roots using this formula.

For example, the cosine of 75 degrees is (using sum of angle formulas) exactly equal to

cos(5 pi/12) = 1/(sqrt(6)+sqrt(2)).

If we don't like the square roots in the denominator, we can multiply both numerator and denominator by (sqrt(6)-sqrt(2)) and the denominator becomes

(sqrt(6))^2-(sqrt(2))^2 = 4.

So another form for the cosine of 75 degrees is

cos(5 pi/12) = (sqrt(6)-sqrt(2))/4.

This is sometimes useful in trying to compute limits.  For example, suppose we seek the limit of

f(x) = (x-4)/(sqrt(x)-2)

as x approaches 4.  If we attempt to substitute in x=4 we simply get an expression which is undefined. But is we multiply numerator and denominator by (sqrt(x)+2), then we get

f(x) = (x-4)(sqrt(x)+2)/(x-4)

which (except at x=4) is equal to

g(x) = sqrt(x)+2.

The limit then, of g(x), as x approaches 4, is equal to g(4) = sqrt(4)+2=2+2=4.  Since g(x) and f(x) agree everywhere except at x=4, they have the same limit there.

Given an expression A+B or A-B, we often speak of the other factor, A-B or A+B, as the conjugate. To simplify the expression f(x) = (x-4)/(sqrt(x)-2), we multiplied by the conjugate of the denominator.

This is especially useful in working with complex numbers. The conjugate of a complex number 

z = a+bi

(where a and b are real numbers) is

z = a–bi.

If we divide one complex number by another, say divide 3+4i by 1-i, we can multiply numerator and denominator by the conjugate of 1-i, that is, multiply by 1+i.

(3+4i)/(1-i) = [(3+4i)(1+i)]/[(1-i)(1+i)].

The denominator will now simplify to 1-i^2 = 2.  So the fraction becomes

(3+4i)/(1-i) = (-1+7i)/2.

The idea of a conjugate appears in the limit of (1-cos x)/x as x approaches zero.  This important limit, used in developing the derivative of the sine function and cosine functions, can be rewritten by multiplying numerator and denominator by the "conjugate" 1+cos x. If we are willing to rewrite 1-cos^2 x as sin^2 x using the trig version of the Pythagorean Theorem, then we have

Once we have this expression, we can recall that the limit of sin x/x is 1 and so

The identity

A^2 – B^2=(A + B)(A – B)

will show up any many areas of mathematics!  Be alert for opportunities to use it!