1. ## Complex Limit (2)

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

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

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

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

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

Now what?

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

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

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

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

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

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

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

if we change to polar coordinates we get

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

Now as $r \to 0$ we get

$2i$

Also, how is sin(rsin(theta)) over rsin(theta) simplifying?

4. Originally Posted by dwsmith

Also, how is sin(rsin(theta)) over rsin(theta) simplifying?
Remember that

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