The Graph of y = ln(x) for positive AND negative values of x.
____________________________________________________________
When we restrict ourselves to the real numbers, ln( – 1 ) does not make sense
because the graph only seems to exist for x > 0
However, if we allow complex y values we can actually find values of
ln(x) for negative x values!
We know the three series for ex, sin(x) and cos(x):
ex = 1 + x + x2 + x3+
2! 3!
sin(x) = x – x3 + x5 – x7
3! 5! 7!
cos(x) = 1 – x2 + x4 – x6
2! 4! 6!
So let us consider e iθ
e iθ = 1 + iθ + (iθ)2 + (iθ)3 + (iθ)4 + (iθ)5 + (iθ)6 + …
2!
3!
4!
5!
6!
= 1 + iθ – θ2 – iθ 3 + θ4 + iθ5 – θ6 – iθ7 + …
2!
3!
4!
5!
6!
7!
= 1 – θ2 + θ4 – θ6 +… + i θ – θ 3 + θ5 – θ7 +…
2!
4!
6!
3!
5!
7!
=
cos(θ)
Or cis(θ) = e iθ
If z = rcis(θ) = reiθ
Then ln (z) = ln(reiθ)
= ln(r) + ln (eiθ)
So
ln(z) = ln(r) + iθ
+ i sin(θ)
Now this is very interesting because if we let z = 1 + 0i
then r = 1 and θ is not just 0 but 2nπ where n∈ Integers
so ln(1) = 0 + 2nπi
This means that for the graph y = ln(x)
if x = 1 then y could be 0 or ±2πi or ±4πi or ±6πi etc
Also if we let z = – 1 + 0i
then r = +1 and θ = π
so ln(– 1) = 0 + (2n + 1)πi
This means that for the graph y = ln(x)
if x = – 1 then y could be ±πi or ±3πi or ±5πi etc
imag z
1
real z
imag z
1
real z
Somehow, the graph of y = ln(x) is not just the RED graph below but it also
passes through all the marked points!
(1, -4πi)
(-1, -3πi)
(1, -2πi)
(-1, -πi)
(1, 0)
(-1, πi)
(1, 2πi)
(-1, 3πi)
(1, 4πi)
If z is just any positive real number z = x + 0i
imag z
x
Real z
then r = x and θ = 0, ±2π, ±4π etc
This means the graph of y = ln(x) really consists of countless parallel graphs
spaced at distances of 2πi to each other.
The equations are y = ln(x) + 2nπi
The RED graph is the usual
y = ln(x)
It is in the x, y plane and
when x = 1 then y = 0
The spacing lines in the
imaginary y direction are at
intervals of 2π rads
The PURPLE graphs are all
parallel to the red graph at
distances of ±2π and ±4π
When x = 1 then y = ± 2πi
and ± 4πi
The equations are:
y = ln(x) ± 2πi and
y = ln(x) ± 4πi
If z is a negative real number z = – x + 0i
then r = │x│ and θ = ±π, ±3π, ±5π etc
imag z
x
Real z
This means the graph of y = ln(x) where x is a negative real number, really
consists of countless parallel graphs also spaced at distances of 2π to each other.
The equations are y = ln│x│ + (2n + 1)πi
The two GREEN
graphs are:
y = ln│x│ ± πi
When x = – 1
then y = ± πi
The two BLUE
graphs are:
y = ln│x│ ± 3πi
When x = – 1
then y = ± 3πi
For better clarity, I will just include y = ln(x) ± 2πi for positive x values
and for negative x values I will just include y = ln│x│ ± πi
This point is ( – 1, πi)
which shows ln(– 1) = πi
This point is (1, 2πi)
which shows ln(1) = 2πi