Complex numbers limits

Question

$\lim_\limits{z\to\infty} \sqrt{z-2i} - \sqrt{z-i} ,$ where z is complex no.

How to evaluate this?

I tried by assuming $z = x+iy$ and evaluated $z-2i = x+ i(y-2)$ and $z-1 = x + i(y-1)$ and after putting the value in the given question , I couldn't think of the next step at all

Writing $\sqrt w$ where $w$ is a complex number is dangerous: complex numbers have two square roots. — Angina Seng, Jul 11 '20 at 16:55
Please see our Help > How can I format mathematics here? about using MathJax/LaTeX for formatting maths. — CiaPan, Jul 11 '20 at 18:58

score 0 · Accepted Answer · answered Jul 11 '20 at 18:34

Typically, to find the limit of $\sqrt{a}- \sqrt{b}$ where both a and b go to infinity, one multiplies by the "unit fraction" $\frac{\sqrt{a}+ \sqrt{b}}{\sqrt{a}+ \sqrt{b}}$. Since $(\sqrt{a}- \sqrt{b})(\sqrt{a}+ \sqrt{b})= a- b$ the fraction becomes $\frac{a- b}{\sqrt{a}+ \sqrt{b}}$.

In this problem, $a= z- 2i$ and $b= z+ i$ so $a- b= z- 2i- z- i= -3i$. The fraction is now $\frac{3i}{\sqrt{z- 2i}+ \sqrt{z+ i}}$. As z goes to infinity, the denominator goes to infinity and, since the numerator is constant, the fraction goes to 0.

But first we need to check if the limit exists or not. How to check that? — user268439, Jul 11 '20 at 19:01

Complex numbers limits

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