Download 2.4 Other complex functions

20
2.4
Other complex functions
Logarithm:
real values: f (x) = ex
f(x)
3
e
x
f(x)=x
2
1
ln x
0
-4
-3
-2
-1
0
1
2
3
x
4
-1
-2
-3
w = ex → x
=
⇒ f −1 (x) = ln x is the inverse function with respect to ex
ln w
ln 1 = 0
ln0 = −∞
ln e = 1
now complex:
w = ez
ea · eib ← 2π periodic in b
⇔ z = ln w = ln |w| + iϕ + 2πki, k ∈ Z
=
test:
ez = eln |w|+iϕ+2πki = eln |w| · ei(ϕ+2πk) = |w|eiϕ = w ⇒ o.k.
Definition 9
f (z) = ln |z| + iϕ + 2πki, k ∈ Z z = |z|eiϕ
is the complex logarithm of z. For k = 0 we get the main value.
Example:
ln(−1)
=
ln 1 +i ϕ +2πki = iπ + 2πki = iπ(2k + 1)
|{z}
|{z}
0
→
π
logarithm of negative numbers now defined
ln 0 still −∞
Exponential function with arbitrary base:
real numbers: ab = eb ln a
a>0
generalization for complex numbers: b ln a
b ∈ C, a ∈ C\{0} ⇒ ab = eb ln a = eb(ln |a|+iϕa +2πik)
Definition 10 00 = 1
special function:
f (z)
f (z)
=
=
az ,
a>0
(a = e-function)
z ln a
=e
=
zb, b ∈ R
e
ln a Re(z) i ln a Im(z)
e
o.k.
important (for root finding):
f (z)
→ f (z)
=
=
different cases: b ∈ Z
z = |z|eiϕ
b(ln |z|+iϕ+2πik)
e
b ln z
e
b(ln |z|+iϕ) 2πikb
→ e
=e
e
2πikb
;
k∈Z
= 1 → o.k. ”normal” power of z
b irrational → e2πikb infinite number of values
b rational → e2πikb =
ˆ finite number of values
2.4 Other complex functions
21
Example:
k
1
: e2πi n k = 0, . . . , n − 1
n
ϕ
1
k
= |z| n ei n e2πi n k = 0, . . . , n − 1
Im{z}1,5
b =
1
⇒ zn
1,0
0,5
0,0
-1,5
-1,0
-0,5
0,0
0,5
1,0
1,5
Re{z}
-0,5
-1,0
-1,5
in particular:
n=2 :
√
1+i
for k = 0
for k = 1
=
:
:
√
√
√
Im{z}
1
2
2 e
1
2
2 e
1
2
2 e
iπ
4
iπ
8
1
2
e
k
2πi n
e0 =
iπ
8 πi
e
√
, k = 0, 1
1
π
2 2 ei 8
√ 12 πi(1+ 1 ) √ 21 9 iπ
8
= 2 e
= 2 e8
1,0
Zi
0,5
0,0
0,0
-0,5
0,5
1,0
Re{z}
1+i
-0,5
⇒ Roots are difficult functions, in particular not single values.
The most simple case we know already from square roots of pure positive real numbers a. The solution is
√
2πik √
a
± a=e 2
which for k ∈ Z has the two independent solutions for k = 0 and k = 1.
Fundamental theorem of algebra
Each polynomial equation of degree n ∈ N
an z n + an−1 z n−1 + . . . + a1 z + a0 = 0
with a0 , . . . , an ∈ C has at least one solution z ∈ C.
⇒ Say this is z1 , then:
(z − z1 ) · bn−1 z n−1 + . . . + b0 = 0
k
⇒ each polynomial has exactly n solutions where so-called ”multiple zeros”, i.e. function (z − z0 ) count k-times.
n=1
n=2
n=3
n=4
n>4
a1 z + a0 = 0
a2 z 2 + a1 z + a0 = 0
a3 z 3 + a2 z 2 + a1 z + a0 = 0
a4 z 4 + . . . + a0 = 0
⇒
⇒
⇒
⇒
⇒
z = − aa10
quadratic equation
solution via Cardano formula
solution via formula possible
no formula exist for a general treatment!