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I tried to simplify it

$$f(z)=\frac{3z^4 -2z^3 +8z^2 -2z +5}{z-i}$$ $$f(z)=\frac{{3z^2 -2z +5}{z^2+1}}{z-i}$$ $$f(z)=\frac{(3z^2 -2z +5)(z+i)(z-i)}{z-i}$$

$$f(z)=(3z^2-2z+5)(z+i)$$

How to prove that it won't be continuous at $i$ from here on?

yesiam
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  • $$3z^4 + 8z^2 + 5 - 2z(z^2 + 1) = (z^2 + 1)(3z^2 + 5) - 2z(z^2 + 1) = (z-i)(z+i)(3z^2 - 2z + 5) = (z-i)(z+i)(3z - 5)(z+1)$$ So, we can write $f(x) = (z+i)(3z - 5)(z+1)$ and therefore, I think it is continuous everywhere. – Snowball Jun 18 '21 at 04:25
  • @Snowball The question says it is supposed to be discontinuous at z=i but putting that in f(z) after simplifying doesn't give 0. Even when applying limits I can't understand how to prove it. – yesiam Jun 18 '21 at 04:35
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    Is $f$ defined at $z=i$? – ultralegend5385 Jun 18 '21 at 04:50
  • It's not continuous at $z=i$, but it's nevertheless weird to say that it's discontinuous there. See here. – Hans Lundmark Jun 18 '21 at 05:48

2 Answers2

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$$f(z) = \left\{ \begin{aligned} (z+i)(3z-5)(z+1), \quad z & \ne i \\ \text{undefined}, \quad z & =i \end{aligned} \right.$$

Since $f(z)$ is undefined at $z=i$, it cannot be continuous there.

It is, though, a removable discontinuity. Just define it to be $\lim_{z\to i} f(z).$

mjw
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  • can you please explain how it is undefined at z=i ? – yesiam Jun 18 '21 at 04:58
  • @yesiam That's because you can't put $z=i$ in $\frac{3z^4 -2z^3 +8z^2 -2z +5}{z-i}$. – azif00 Jun 18 '21 at 05:06
  • @mjw So to check if a function is defined at a point we put the point in the initial equation and not the simplified one? Sorry I'm a bit rusty with the concepts. – yesiam Jun 18 '21 at 05:12
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    Well, the definition of $f(z)$ is a fraction where the numerator and denominator are both zero at $z=i$. Because the denominator is zero, the function is not well defined there. Of course, this fraction has a finite limit, and we can define $f(z)$ to be the value of the limit, but we need to define it to remove the 'singularity'. It's a subtle point, but I think that is what the question is asking. – mjw Jun 18 '21 at 05:25
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Recall the limit definition of continuity.

A function $f$ is continuous at $x=c$ iff $$\lim_{x\to c}f(x)=f(c)$$

When $f(c)$ is not defined, it doesn't make sense to compare it to the limit. So, the function is not continuous. $\square $

However, it is a removable discontinuity as we can define (this function is different from the original function) $f(z)$ to be the original fraction for $x\neq i$ and $f(i)$ as the limit of $f(z)$ as $z\to i$.

Also, kudos to you to post the problem with context and details. A great welcome to MathSE!

Hope this helps. Ask anything if not clear :)

ultralegend5385
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