5.4 Complex conjugate

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DEFINITION 5.4.1 (Complex conjugate) If z = x + iy, the complex

conjugate of z is the complex number de¯ned by z = x ¡ iy. Geometrically,

the complex conjugate of z is obtained by re°ecting z in the real axis (see

Figure 5.2).

The following properties of the complex conjugate are easy to verify:

5.4. COMPLEX CONJUGATE 97

¾ -

6

?

x

y

z

z

½

½

½

½>

Z

Z

Z

Z~

Figure 5.2: The complex conjugate of z: z.

1. z1 + z2 = z1 + z2;

2. ¡z = ¡z.

3. z1 ¡ z2 = z1 ¡ z2;

4. z1z2 = z1 z2;

5. (1=z) = 1=z;

6. (z1=z2) = z1=z2;

7. z is real if and only if z = z;

8. With the standard convention that the real and imaginary parts are

denoted by Re z and Im z, we have

Re z =

z + z

2

; Im z =

z ¡ z

2i

;

9. If z = x + iy, then zz = x2 + y2.

THEOREM 5.4.1 If f(z) is a polynomial with real coe±cients, then its

non{real roots occur in complex{conjugate pairs, i.e. if f(z) = 0, then

f(z) = 0.

Proof. Suppose f(z) = anzn + an¡1zn¡1 + ¢ ¢ ¢ + a1z + a0 = 0, where

an; : : : ; a0 are real. Then

0 = 0 = f(z) = anzn + an¡1zn¡1 + ¢ ¢ ¢ + a1z + a0

= an zn + an¡1 zn¡1 + ¢ ¢ ¢ + a1 z + a0

= anzn + an¡1zn¡1 + ¢ ¢ ¢ + a1z + a0

= f(z):

98 CHAPTER 5. COMPLEX NUMBERS

EXAMPLE 5.4.1 Discuss the position of the roots of the equation

z4 = ¡1

in the complex plane.

Solution. The equation z4 = ¡1 has real coe±cients and so its roots come

in complex conjugate pairs. Also if z is a root, so is ¡z. Also there are

clearly no real roots and no imaginary roots. So there must be one root w

in the ¯rst quadrant, with all remaining roots being given by w; ¡w and

¡w. In fact, as we shall soon see, the roots lie evenly spaced on the unit

circle.

The following theorem is useful in deciding if a polynomial f(z) has a

multiple root a; that is if (z ¡a)m divides f(z) for some m ¸ 2. (The proof

is left as an exercise.)

THEOREM 5.4.2 If f(z) = (z ¡ a)mg(z), where m ¸ 2 and g(z) is a

polynomial, then f0(a) = 0 and the polynomial and its derivative have a

common root.

From theorem 5.4.1 we obtain a result which is very useful in the explicit

integration of rational functions (i.e. ratios of polynomials) with real coe±-

cients.

THEOREM 5.4.3 If f(z) is a non{constant polynomial with real coe±-

cients, then f(z) can be factorized as a product of real linear factors and

real quadratic factors.

Proof. In general f(z) will have r real roots z1; : : : ; zr and 2s non{real

roots zr+1; zr+1; : : : ; zr+s; zr+s, occurring in complex{conjugate pairs by

theorem 5.4.1. Then if an is the coe±cient of highest degree in f(z), we

have the factorization

f(z) = an(z ¡ z1) ¢ ¢ ¢ (z ¡ zr) £

£(z ¡ zr+1)(z ¡ zr+1) ¢ ¢ ¢ (z ¡ zr+s)(z ¡ zr+s):

We then use the following identity for j = r + 1; : : : ; r + s which in turn

shows that paired terms give rise to real quadratic factors:

(z ¡ zj)(z ¡ zj) = z2 ¡ (zj + zj)z + zjzj

= z2 ¡ 2Re zj + (x2

j + y2

j );

where zj = xj + iyj .

A well{known example of such a factorization is the following:

5.5. MODULUS OF A COMPLEX NUMBER 99

¾ -

6

?

jzj

x

y

z

½

½

½

½>

Figure 5.3: The modulus of z: jzj.

EXAMPLE 5.4.2 Find a factorization of z4+1 into real linear and quadratic

factors.

Solution. Clearly there are no real roots. Also we have the preliminary

factorization z4 + 1 = (z2 ¡ i)(z2 + i). Now the roots of z2 ¡ i are easily

veri¯ed to be §(1 + i)=p2, so the roots of z2 + i must be §(1 ¡ i)=p2.

In other words the roots are w = (1 + i)=p2 and w; ¡w; ¡w. Grouping

conjugate{complex terms gives the factorization

z4 + 1 = (z ¡ w)(z ¡ w)(z + w)(z + w)

= (z2 ¡ 2zRew + ww)(z2 + 2zRew + ww)

= (z2 ¡

p2z + 1)(z2 + p2z + 1):

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