Complex Functions and Analyticity - (4.1)
1. Real and Imaginary Parts of Complex Functions
Consider complex functions fz : ℂ → ℂ. These functions are transformations (mappings) of the
complex plane into itself. For example, fz  z n , fz  e z , fz  z . Let z  x  iy. For each complex
1−z
function fz, there are associated two real-valued functions ux, y and vx, y such that
fz  ux, y  ivx, y
where ux, y and vx, y are real-valued functions.
Example Find the real and imaginary parts of fz  z 4 . Evaluate f2 − 3i.
fz  z 4  x  iy 4  x 4  4x 3 iy  6x 2 iy 2  4xiy 3  iy 4
 x 4 − 6x 2 y 2  y 4  i4x 3 y − 4xy 3 
Hence, ux, y  x 4 − 6x 2 y 2  y 4 and vx, y  4x 3 y − 4xy 3 .
f2 − 3i  u2, −3  iv2, −3  2 4 − 62 2 −3 2  −3 4  i 42 2 −3 − 42−3 3
 − 34 499  168i
Example Find the real and imaginary parts of fz  e .
z
fz  e z  e xiy  e x cosy  i siny, ux, y  e x cosy, vx, y  e x siny.
z . Evaluate f1  i
1−z
x1 − x − y 2
y1 − x  xy
1 − x  iy

i
2
2
1 − x  iy
1 − x  y
1 − x 2  y 2
Example Find the real and imaginary parts of fz 
fz 
z  x  iy
1−z
1 − x − iy
x1 − x − y 2
y1 − x  xy
and vx, y 
.
2
2
1 − x  y
1 − x 2  y 2
11 − 1  11
11 − 1 − 1 2

i
 −1i
f1  i 
2
1 − 1  1 2
1 − 1 2  1 2
Hence, ux, y 
Example Find the real and imaginary parts of fz  sin2z. Evaluate f 3  4 i .
iaz
−iaz
iaz
−iaz
By Euler formula, we have sinaz  e − e
and cosaz  e  e
. Hence,
i2
2
iaxiy
− e −iaxiy  e iax−ay − e −iaxay
sinaz  e
i2
i2
i
−ay
 − e cosax  i sinax − e ay cosax − i sinax
2
1
 −ie −ay cosax  e −ay sinax  ie ay cosax  e ay sinax
2
 1 e −ay sinax  e ay sinax  i 1 e ay cosax − e −ay cosax
2
2
 coshay sinax  i sinhay cosax
Note that sinaz  sinax  iay  sinax cosiay  siniay cosax. Hence,
cosiay  coshay, siniay  i sinhay
sin 2   i
2
3
1
 cosh 
2
sin 2
3
 i sinh 
2
cos 2
3
Example Find the real and imaginary parts of fz Lnz. Evaluate f1  2i.
Lnz Ln|z|e iArgz   ln|z|  lne iArgz   ln|z|  iArgz  lnr  i, −    ≤ .
y
ux, y  ln x 2  y 2  12 lnx 2  y 2 , vx, y  tan −1 x .
2. Regions in Complex Plane:
Special regions:
(1) D r  z ∈ ℂ; |z| ≤ r
(2) -neighborhood of z 0 : N  z 0   z ∈ ℂ; |z − z 0 |  
(3) Complement of a region U: U  ℂ\U.
A region U in ℂ is open if for each point z 0 in U there exists a -neighborhood N  z 0  such that
N  z 0  ⊂ U. A region is closed if its complement is open. A region is bounded if it is contained in D r for
some r. A region is said to be compact if it is bounded and closed.
For example, N 0.5 1  i  z ∈ ℂ; |z − 1  i|  0. 5 . Notice that N 0.5 1  i is not closed since all z
on the boundary do not belong to N 0.5 1  i but N 0.5 1  i is closed. Graphically,
y
1.5
y
1.5
1.0
1.0
0.5
0.5
0.0
0.0
0.0
0.5
1.0
0.0
1.5
x
0.5
1.0
1.5
x
N 0.5 : all z inside the circle
N  z 0  in N 0.5 1  i
A region U is connected if any two points of U can be connected by a finite sequence of line segments
lying entirely in U. U is simply connected if it is connected and any simple closed curve can be shrunk to a
point continuously in U. For example,
y
1.5
y
1.5
y
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.0
0.0
0.0
0.5
1.0
not connected
1.5
x
0.0
0.0
0.5
1.0
connected but not simply
1.5
x
0.0
0.5
1.0
connected
(3) Limit and Continuity:
Let fz  uz  ivz. Then
(i) lim z→z 0 fz  lim z→z 0 uz  i lim z→z 0 vz.
(ii) fz is continuous at z 0  x 0  iy 0 if and only if ux, y and vx, y are continuous at x 0 , y 0 .
2
1.5
x
Example Compute lim z→aib e z . Determine if e z is continuous everywhere.
lim e z 
z→aib
lim
x,y→a,b
e x cosy  i siny 
lim
x,y→a,b
e x cosy  i
lim
x,y→a,b
e x siny
 e a cosb  ie a sinb
Because ux, y  e x cosy and vx, y  e x siny are continuous everywhere, e z continuous everywhere.
Example Determine where (1) fz  z4  1 is not continuous.
z 1
fz is not continuous when z 4  1  0, z 4  −1  e i , z  4 −1  e i


1
z 1  e i 4   12  12 i 2 , z 2  e i 4 2 4  −  12 − 12 i 2


2
3
z 3  e i 4 2 4  −  12  12 i 2 , z 4  e i 4 2 4   12 − 12 i 2

4
2 4k
, k  0, 1, 2, 3
4 Complex Polynomials:
Let fz in ℂz, say fz  a n z n  a n−1 z n−1 . . . a 1 z  a 0 . Then
(i) fz is continuous everywhere in ℂ.
(ii) lim z→ |fz|   if fz is not a constant. (If |fz| ≤ K for some K, then fz  C. )
(iii) |fz| is bounded on every compact region in ℂ. (Extreme Value Theorem)
5. Derivatives:
fz  z − fz
.
z
All differentiation rules: power rule, sum rule, product rule, quotient rule, chain rule, apply here.
For example,
d z n   nz n−1
dz
d a n z n  a n−1 z n−1  a 1 z  a 0   na n z n  n − 1a n−1 z n−1 . . . 2a 2 z  a 1
dz
d n a n z n  a n−1 z n−1  a 1 z  a 0   a n n!, d n a n z n  a n−1 z n−1  a 1 z  a 0   0
dz n
dz n
′
f z  lim
z→0
6. Analytic Functions:
The function fz is analytic at z 0 if fz is differentiable in a circular neighborhood of z 0 . fz is analytic
in U if it is analytic every point in U. fz is called an entire function if it is analytic at every point in ℂ.
For example, polynomials are entire functions.
3