prove that Limit of $sin\bar{z} \over{sinz} $ does not exists as $z\to 0$

Question

so i have to prove that the limit of this function doenot exists

$sin\bar{z} \over{sinz} $ as $z\to 0$

well i started with z = $re^{i\theta}$

so

${sin\space r\space e^{-i\theta}} \over sin \space r \space e^{i\theta}$

so $\lim z \space \space\space when \space\space z\to0 $

$ sin \space e^{-2i\theta} \space \space\space\space is \space depends \space on \space \theta \space only \space so \space lim \space doesn't \space exists \space for\space this \space function \space right \space ?? $

Answer 1

If $z = x \in \mathbb R$, then $$ \frac{\sin \overline z}{\sin z} = \frac {\sin x}{\sin x} = 1 \to 1; $$ if $z = \mathrm i y, y \in \mathbb R$, then $$ \frac {\sin \overline z}{\sin z} = \frac {\sin(-\mathrm i y)}{\sin (\mathrm i y)}, $$ using the series expansion of $\sin z$ we have $\sin (-z) = -\sin(z)$, so the quotient above equals $-1$.

Thus the limit does not exist.

Answer 2

From $\lim_{z\to 0}\frac{z}{\sin z}=1$ we have that

$$\lim_{z\to 0}\frac{\sin\bar z}{\sin z}=\lim_{z\to 0}\frac{\bar z}{z}=\lim_{r\to 0^+}\frac{re^{-i\theta}}{r e^{i\theta}}=e^{-i2\theta}$$

for $z=re^{i\theta}$. Hence the limit doesn't exists because for different $\theta\in [0,\pi)$ the value $e^{-i2\theta}$ changes.