If $z=(3+7i)(p+iq)$, where $p,q \in Z-{\{0\}}$ is purely imaginary, then the minimum value for $|z|^2$ is

Question

Options

A) 0

B) 58

C) 3364

I know this question has already been asked on this site, but I have a different problem with it

I did the obvious, and obtained $$3p-7q=0$$

For the minimum value of $|z|^2$, the answer will be $(Im(z))^2$

$$=(3q+7p)^2$$

Putting $q =\frac 37 p$

$$=(\frac 97 p +7p)^2$$ $$=p^2(\frac{58}{7})^2$$

Since there are no restrictions on $p$ (other than that it cannot be 0), I could put value as $\frac{7}{\sqrt {58}}$ to get $58$ as the answer

Now, the right answer is $3364$, and I know all the process behind it, but can I get an explanation for this contradiction?

Answer 1

There are restrictions. Notice that they belong to $Z$, the family of INTEGERS. That's why it must be $58^2$.

Answer 2

For $3p-7q=0$ and $p$ and $q$ to both be integers, we need that $7$ divides $p$. So, we can write $p=7m$ for some (nonzero) integer $m$, and then $$|z|^2=58^2m^2.$$