What is $\lim_{z \to \infty} \frac{z^{2}-(2+3i)z+1}{iz-3}$?

Question

What is $$\lim_{z \to \infty} \frac{z^{2}-(2+3i)z+1}{iz-3}\;?$$

I know the limit comes out to be infinity. But does having a complex number make any difference in the answer? Moreover, does this limit even exist?

Answer 1

I do not know about complex analysis and what $z \rightarrow \infty$ means. But, I guess if $z \rightarrow \infty \Rightarrow |z| \rightarrow \infty$. Using this I show that $\lim_{|z| \rightarrow \infty}\mid \frac{z^2 -(2+3i)z +1}{iz-3} \mid \rightarrow \infty$.

I guess this could mean, $\lim_{z \rightarrow \infty} \frac{z^2 -(2+3i)z +1}{iz-3} \rightarrow \infty$.

Proof:

Let, $arg(z) = \theta$ and let, $z' = iz + 3$,

It can be shown that $ |z'| = \sqrt{|z|^2 - 6|z|sin\theta + 9 }$, since $|z| \rightarrow \infty$, $|z'| \lt 2|z|$

$$\Rightarrow \mid \frac{z^2 -(2+3i)z +1}{iz-3} \mid = \mid \frac{z^2 -(2+3i)z +1}{z'} \mid \gt \mid \frac{z^2 -(2+3i)z +1}{2z} \mid $$

$$\Rightarrow \mid \frac{z^2 -(2+3i)z +1}{iz-3} \mid \gt \frac{1}{2} \big(|z| - \sqrt{13} - \frac{1}{|z|} \big)$$

$$\therefore \lim_{|z| \rightarrow \infty} \mid \frac{z^2 -(2+3i)z +1}{iz-3} \mid \gt \lim_{|z| \rightarrow \infty} \frac{1}{2} \big(|z| - \sqrt{13} - \frac{1}{|z|} \big)$$

$$\Rightarrow \lim_{|z| \rightarrow \infty} \mid \frac{z^2 -(2+3i)z +1}{iz-3} \mid \rightarrow \infty$$