Calculator test

[Commie]

Hi Chaps,

The correct equation for Ln(z)=Ln(x+iy) is:
\[ln({x+iy}) = \frac{1}{2}ln({x^{2}+y^{2}})+{i}arctan(\frac{y}{x}) \pm2k\pi i\]

Consider, you have z=3-4i. we have Ln(3-4i)=1.6094-0.9273i+0i , k=0

We can take e^(1.6094-0.9273i) = 3-4i back where we started.

If we now increment/decrement it's imaginary part by 2.pi then

Ln(3-4i)=1.6094-0.9273i+2.pi.i= 1.6094+5.3559i , k=1
we can take e^(1.6094+5.3559i)= 3-4i back where we started.

Now increment/decrement it's imaginary part by another 2.pi then

Ln(3-4i)=1.6094-0.9273i+4.pi.i= 1.6094+11.6391i k=2
We can take e^(1.6094+11.6391i)= 3-4i back where we started.

When k=0, the imaginary is in ground state, i.e., the smallest possible(principal) value for the imaginary i.

And so on, these imaginary 2.pi jumps exist between +- infinity. However, no matter how large the imaginary grows in 2.pi increments or decrements, it will always yield, in the example shown, (3-4i) back where we started.

The real always stays constant, whilst the imaginary i is quantized in steps of 2.pi.

Stay safe guys
Darren