we can know that 1^((-i*ln2)/(2π))=2, but why log₁2 is still undefined?
logarithms - log base 1 of 1 - Mathematics Stack Exchange
Can somebody please explain Logarithm to me at basic level?
ELI5: What is a natural logarithm? Why is ln(1) = 0?
complex analysis - How to derive the value of $\log(-1)$? - Mathematics Stack Exchange
What is the general logarithm formula for log 1?
What is the value of log10 1?
What are common mistakes when calculating log 1?
Videos
The reason why it is not convenient to define for the base of
is simple:
But the denominator is and thus the division doesn't make any sense unless we're working with limits :)
What is ? What number must
be if
? It can be any number.
What is ? What number must
be if
? It can be any number.
Hence these expressions are undefined.
What is if
? In some cases it's
. It depends on which functions
and
are. It can be any number or
or
. But it's not always undefined. In many cases it's defined and equal to a particular number. For that reason
is an indeterminate form.
What is if
? Again this depends on which functions
and
are. In many cases it's a specific number. This is also an indeterminate form.
Say I want to Find Log(10)1.5, I know that Log1=0 and Log2=0.3010, how can I use this relation to find the log of 1.5 with base 10?
Hint
Using Euler identity, for integer $n$, $$\cos ( (2 n+1)\pi)+i \sin ( (2 n+1)\pi)=e^{i(2 n+1)\pi}=-1$$
On the complex plane, $e^z = e^{z+2\pi ki}$ for any integer $k$. This is a bit funny, right? While the equality holds, we see that $z \ne z + 2\pi ki$ ! So if we try the inverse of the exponential function, it's a bit ambiguous as to what number we get back. Does $\log e^z = z$ or does $\log e^z = z + 2\pi i$?
The logarithm can return different complex numbers! It appears that the complex logarithm, is multi-valued. Here's a common definition:
$$\log z = \log |z| + i \arg z +i2\pi k$$
Where $\arg z$ returns the argument or angle of the complex number $z$. Note that the final term $i2\pi k$ accounts for the infinitely many values that can be returned. If we let $z=re^{i\phi}$ and substitute into the line above, we get
$$\log z = \log r + i\phi +i2\pi k$$
Then
$$\log (-1) = \log |-1| + i(2k+1)\pi = i(2k+1)\pi$$
Now this multi-valued business can be inconvenient, so we restrict the outputted values. This is called choosing a branch. A common, or principal branch is where $\phi \in [0,2\pi)$. Another common one is where $\phi \in [-\pi,\pi)$. Note that the branch we choose covers the plane entirely, and turns the multivalued function into a single valued one. Choosing the branch $\phi \in [0,2\pi)$, we can then say
$$\log_0 z = \log r + i\phi$$
where I use a subscript $0$ to denote my chosen branch. Then
$$\log_0( -1) = \log |-1| + i\pi = i\pi$$