Trouble understanding $\lim_{z \to 0} \frac{1}{z} = \infty$ in complex

Question

Is it true, that $$\lim_{z \to 0} \frac{1}{z} = \infty$$ for $z \in \mathbb{C}$?

I have some trouble with a definition of this. I know that in $\mathbb{C}$ we have only one point at infinity unlike in $\mathbb{R}$ where we have $\pm \infty$. Could somebody provide me with a definition of the above limit?

Yes - $\lim_{z\to z_0}f(z) = \infty$ if for every $M>0$, there is a $\delta>0$ such that $|f(z)|>M$, whenever $0<|z-z_0|<\delta$. In your example, simply take $\delta = 1/M$. — Mark McClure, Apr 14 '17 at 19:13
It's not really a definition but maybe it will help you understand the definition, we say that $\lim_{z\to a} f(z) = \infty$ if $\lim_{z\to a} |f(z)| = + \infty$. — , Apr 14 '17 at 19:19
@markmcclure Apparently those named Mark think alike. I was wrote my solution and just noticed thereafter that your comment. -Mark — Mark Viola, Apr 14 '17 at 19:27

score 4 · Accepted Answer · answered Apr 14 '17 at 19:14

4

The limit $\lim_{z\to 0}\frac1z =\infty$ means that for all $B>0$, there exists a $\delta>0$ such that $$\left|\frac{1}{z}\right|>B$$whenever $0<|z|<\delta$.

With $\delta=1/B$, we are done.

answered Apr 14 '17 at 19:14

Mark Viola

179,405

score 3 · Answer 2 · answered Apr 14 '17 at 19:20

3

If $z = x+iy$, $\frac1{z} =\frac1{x+iy} =\frac{x-iy}{x^2+y^2} =\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2} $.

If $y = \sqrt{x}$, then, as $x \to 0$, $\frac1{z} =\frac{x}{x^2+x}-i\frac{\sqrt{x}}{x^2+x} \approx 1-i\frac{1}{\sqrt{x}} $, so the real part is bounded while the imaginary part blows up.

Similarly, if $x = \sqrt{y}$, the imaginary part is bounded while the real part blows up.

In all cases, the magnitude of $\frac1{z}$ blows up as $|z| \to 0$.

answered Apr 14 '17 at 19:20

marty cohen

107,799

1

I was going to write something tantamount to this, but thought I'd stick to the definition. (+1) my friend. – Mark Viola Apr 14 '17 at 19:25
Thanks. I hadn't looked at it in this way before, so I found it interesting that the real part or the imaginary part can be bounded. – marty cohen Apr 14 '17 at 20:27
Indeed. Complex analysis is rich with interesting "stuff." – Mark Viola Apr 14 '17 at 20:38

Trouble understanding $\lim_{z \to 0} \frac{1}{z} = \infty$ in complex

2 Answers2