# Thread: natural log and imaginary unit question

1. ## natural log and imaginary unit question

Hi,

What happens when you take the natural log of a number that involves i?

for eg.:

$\ln{ix}$

or even better:

$i\ln{ix}$

Or, put another way, what happens when you take the integral of a number involving i?

$\ln{ix}=\int_{1}^{x}\frac{1}{ix}dx$

Thanks

2. Originally Posted by rainer
What happens when you take the natural log of a number that involves i?
First it is $\log(z)=\log(x+yi)$.
The definition of complex logarithm function is $\log(z)=\ln(|z|)+i\cdot \arg(z)$.

3. I think you misplaced a parenthesis but I got it anyway. Thanks.

4. Originally Posted by rainer
I think you misplaced a parenthesis but I got it anyway. Thanks.
There is no missplaced parenthesis.
$\log(1-i)=\ln(\sqrt2)+i\left(\frac{-\pi}{4}+2k\pi\right)$

5. Ok, so if you didn't misplace a parenthesis I'm still confused.

I'll just cut to the chase. I need help with this:

$\ln{i(\cosh1+\sqrt{\left|1-\cosh1\right|})}$

Thanks

6. The principal value of $\log\left(\cosh(1)+\sqrt{|1-\cosh(1)|}\right)= \ln\left(\cosh(1)+\sqrt{|1-\cosh(1)|}\right)+\frac{i\pi}{2}$.