Download MATH20142: COMPLEX ANALYSIS 1. Construction and basic

MATH20142: COMPLEX ANALYSIS
1. Construction and basic properties of complex numbers
Construction. Given ax2 + bx + c = 0 (a, b, c ∈ R), solution is
√
−b ± b2 − 4ac
x=
.
2a
No solution in R if b2 − 4ac < 0.
√
Euler (1707-1783) introduced the symbol i = −1, so i2 = −1. Then i is a root of
x2 + 1 = 0.
An expression a+ib (a, b ∈ R) is called a complex number. We denote the set of complex
numbers by C.
We add them according to the rule
(a + ib) + (c + id) = (a + c) + i(b + d)
and multiply them according to the rule
(a + ib)(c + id) = ac + iad + ibc + i2 bd = ac + iad + ibc − bd = (ac − bd) + i(ad + bc).
More formally, we can define complex numbers as pairs (a, b) of real numbers with
addition and multiplication
(a, b) + (c, d) = (a + c, b + d),
(a, b)(c, d) = (ac − bd, ad + bc)
and the rule −(a, b) = (−a, −b).
To divide complex numbers, we use the trick
1
a − ib
a − ib
a − ib
a
b
=
= 2
= 2
= 2
−i 2
.
2
2
2
2
a + ib
(a + ib)(a − ib)
a −i b
a +b
a +b
a + b2
If z = x + iy (common to use z for a complex number), x, y ∈ R, then we call x the real
part of z and y (not iy!!!) the imaginary part of z. We also use the notation x = Rez and
y = Imz.
We say z is real if Imz = 0 and imaginary if Rez = 0.
Exercises. For z, w ∈ C, show
(1) Re(z ± w) = Rez ± Rew
(2) Im(z ± w) = Imz ± Imw
(3) Re(zw) is not always equal to (Rez)(Rew)
(4) Im(zw) is not always equal to (Imz)(Imw)
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If z = x + iy, x, y ∈ R, then x − iy is called the complex conjugate of z, written z. (Used
earlier for division.)
Exercises. For z, w ∈ C, show
(1) z ± w = z ± w
(2) zw = zw
¡ ¢
(3) wz = wz if w 6= 0.
Note that
z + z = 2Rez,
z − z = 2iImz.
Modulus and argument. Let z = x + iy, x, y ∈ R. We visualize z as represented by
the point (x, y) in the plane. (Picture.) This is called the complex plane or the Argand
diagram.
Real numbers are represented by points on the horizontal axis and imaginary numbers
by points on the vertical axis.
The modulus (or absolute value) of z = x + iy ∈ C is
|z| =
p
x2 + y 2 ≥ 0.
If z ∈ R, this is just the usual absolute value.
We have
|z| = |z|
and
zz = (x + iy)(x − iy) = x2 + y 2 = |z|2 .
The angle θ in the figure is called the argument of z,
θ = arg z.
Also
tan θ = y/x.
Note θ is not uniquely determined: if we replace θ by θ + 2nπ, n ∈ Z, then we get the
same point. However there is a unique value of θ such that
−π < θ ≤ π.
This is called the principal value of arg z.
Proposition 1.1. Let z, w ∈ C. Then
(1)
(2)
(3)
(4)
|z| = 0 if and only if z = 0;
|zw| = |z||w|;
¯z¯
¯ ¯ = |z| if w 6= 0;
w
|w|
|z + w| ≤ |z| + |w| (triangle inequality)
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Proof. (i) from figure.
(ii)
|zw|2 = (zw)(zw) = zwzw = (zz)(ww) = |z|2 |w|2
and take square roots.
(iii) exercise.
p
(iv) Note z = x + iy =⇒ |z| = x2 + y 2 ≥ |x|. Now
|z + w|2 = (z + w)(z + w) = (z + w)(z + w)
= zz + ww + zw + wz
= |z|2 + |w|2 + zw + zw
= |z|2 + |w|2 + 2Re(zw)
≤ |z|2 + |w|2 + 2|zw|
by above
= |z|2 + |w|2 + 2|z||w| by (ii)
= |z|2 + |w|2 + 2|z||w|
= (|z| + |w|)2 ,
so |z + w| ≤ |z| + |w|.
¤
Exercise. Show that
|z − w| ≥ ||z| − |w||.
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