1. ## Complex Limit (2)

$\displaystyle \lim_{z\to0}\frac{e^z-e^{\bar{z}}}{Im(z)}$

$\displaystyle e^z=e^x(cos(y)+isin(y))$

$\displaystyle e^{\bar{z}}=e^x(cos(y)-isin(y))$

$\displaystyle \lim_{z\to0}e^x\frac{cos(y)+isin(y)-(cos(y)-isin(y))}{y}$

$\displaystyle \lim_{z\to0}e^x\frac{2isin(y)}{y}$

Now what?

2. Originally Posted by dwsmith
$\displaystyle \lim_{z\to0}\frac{e^z-e^{\bar{z}}}{Im(z)}$

$\displaystyle e^z=e^x(cos(y)+isin(y))$

$\displaystyle e^{\bar{z}}=e^x(cos(y)-isin(y))$

$\displaystyle \lim_{z\to 0}e^x\frac{cos(y)+isin(y)-(cos(y)-isin(y))}{y}$

$\displaystyle \lim_{z\to0}e^x\frac{2isin(y)}{y}$

Now what?
$\displaystyle \lim_{z\to 0 }e^x\frac{2isin(y)}{y}$

$\displaystyle \lim_{(x,y)\to (0,0)}e^x\frac{2isin(y)}{y}$

if we change to polar coordinates we get

$\displaystyle \lim_{r\to 0}e^{r\cos(\theta)}\frac{2isin(r\sin(\theta))}{r\s in(\theta)}$

Now as $\displaystyle r \to 0$ we get

$\displaystyle 2i$

$\displaystyle \lim_{u \to 0}\frac{\sin(au)}{au}=1$ for any real number $\displaystyle a$