Let $f(z)\sim e^{-z^4}$ with $z \in \mathbb{C} $. Now the claim is that $\lim_{z\rightarrow \infty} f(z) =0$ along the line $\theta= \pi/16$.
Let's check: Along the given line, limit reduces to $$\lim_{r \rightarrow \infty} e^{-r^4 (\cos(\pi/4)+i\sin(\pi/4))}=e^{-\infty-i \infty}.$$ Obiously it does not seem to be zero.
Could you all please help me out in this ?
On
your conclusion isn't correct. First, if $z = re^{i \pi/16}$, then $$-z^4 = -r^4 e^{i \pi/4} = -r^4 \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) = -r^4 \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right).$$ Note that you did not distribute the negative sign, but that is largely inconsequential.
Now if we look at the magnitude of $$e^{-z^4} = e^{-r^4 (1+i)/\sqrt{2}} = e^{-r^4/\sqrt{2}} e^{-r^4 i/\sqrt{2}},$$ the second factor always has magnitude $1$ for any $r \in \mathbb R$. The first factor, $e^{-r^4/\sqrt{2}}$, is a real number and it is trivial to see $$\lim_{r \to +\infty} e^{-r^4/\sqrt{2}} = 0.$$ So $$|e^{-z^4}| \to 0.$$
Hint. Note that if $z=re^{i\theta}$ with $r\in\mathbb{R}$ ($z$ is along a line with a given angle $\theta$) then $$\left|e^{-z^4}\right|=\left|e^{-r^4 (\cos(4\theta)+i\sin(4\theta))}\right|=e^{-r^4\cos(4\theta)}.$$ What happens to $\left|e^{-z^4}\right|$, as $|r|$ goes to $+\infty$, when $\cos(4\theta)>0$?