Taking complex solution on $\frac{1}{x^2 + 1} = 0$

Question

Is $i$ a complex solution to $$ \frac{1}{x^2 + 1} = 0 \;?$$
Knowing that it has no real solutions, substituting $i$ leaves $\frac{1}{0}$ which for me is equal to zero. Is this right?

Answer 1

No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $\mathbb C$, it would be :

$$f(z) = \frac{1}{z^2+1}, \quad z \in \mathbb C\setminus\{\pm i\}$$

The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.

The equation

$$f(z) = 0 \implies \frac{1}{z^2+1} = 0$$

is false, as it has no solutions at all.

Answer 2

We have that

$$\frac{1}{x^2 + 1} = 0$$

is an expression defined for $x^2+1 \neq 0$ and it has not (real nor complex) solutions.

Indeed for any $z\in \mathbb{C}$ we have

$$\frac{1}{z^2 + 1} = 0 \implies \left|\frac{1}{z^2 + 1}\right| = |0| $$

but

$$\left|\frac{1}{z^2 + 1}\right|=\frac{1}{|z^2 + 1|}\neq 0\quad \forall z$$

Answer 3

No, the complex field does not provide solution to $$\frac {1}{x^2+1}=0$$

That is equivalent to $$1=0\times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.

Answer 4

Note that for $x\in\mathbb{C}$ and polynomials $P$ and $Q$,

$$ \frac{P(x)}{Q(x)} = 0 \Leftrightarrow P(x) = 0 \text{ and } Q(x)\neq 0 $$

Since in this case $P(x) \equiv 1 \neq 0$, there are no $x$ such that $\frac{P(x)}{Q(x)} = 0$.