Question

Complex numbers,


Is this statement true or false and why ?
https://www.dropbox.com/s/fc4yihjnbouj46p/Screenshot%202013-11-25%2001.13.01.jpg

Answer 1

false, I think it's tangent.

Answer 2

can you please tell me how did you solve it @abb0t ?

Answer 3

I think it's true.
Suppose z = x + iy. We'll need the useful fact that |z|^2 = zz* = (x+iy)(x-iy)
So we've got
\[\frac{z}{z+1} = \frac{x+iy}{x+1+iy} = \frac{(x+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}\]
What's really good about this is we can write it as
\[\frac{(x+iy)(x-iy) + (x+iy)}{|z+1|^{2}} = \frac{|z|^{2} + x + iy}{|z+1|^{2}}\]
So we have
\[e^{\frac{z}{z+1}} = \exp(\frac{|z|^{2}+x}{|z+1|^{2}})\exp(iy)\]
(Using exp to replace e, so you can read it).
The real part of this, using \[e^{ix} = cos(x) + isin(x)\] is the result given.

Answer 4

hmm

Answer 5

there is a mistake
\[\large \cdots=\frac{|z|^2+x+iy}{|z+1|^2}=\frac{|z|^2+x}{|z+1|^2}+
i\frac{y}{|z+1|^2} \]

Answer 6

Oh, yeah, thanks! Sorry, tried to type it too fast.

Answer 7

so
\[\large \exp\left(\frac{z}{z+1}\right)=\exp\left(\frac{|z|+x}{|z+1|^2}\right)\cdot
\exp\left(\frac{iy}{|z+1|^2}\right) \]
and the real part of this is as stated following the last identity of Eigensheep.