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

I tried to solve it by multiplying the 2 equations i.e. $z=3p+3iq+7ip+7i^2q$. Then $z=(3p-7q)+i(7p+3q)$.

Then,$|z^2|=z \bar z$

$\bar z =(3p-7q)-i(7p+3q)$. Also since $z$ is purely imaginary then $(3p-7q)=0$. Thus $p=\frac73q$.

Hence $|z|^2=9p^2+49q^2-42pq+1(9q^2+49p^2+42pq)$.

So $|z|^2=18p^2+98q^2$.


$=196 q^2$. After this I don't how to proceed. The answer is $3364$.

  • 1
    $\begingroup$ $|z^2|=(2^2+7^2)(p^2+q^2)$. $\endgroup$ Aug 4, 2017 at 21:01

2 Answers 2


I believe what you're after is the minimum value of $|z|^2=z\bar{z}$, not the minimum value of $|z^2|=\sqrt{\text{Re}^2(z^2)+\text{Im}^2(z^2)}$. I will assume that this is the case.

As you mentioned in your original post, $z=3p+7ip+3iq-7q$. If $p,q\in\mathbb{I}\backslash\{0\}$, then $\text{Re}(z)=7ip+3iq$ and $\text{Im}(z)=3p-7q$, whereby $\bar{z}=7ip+3iq-3p+7q$. I'll let you carry on from here.

It seems odd that $p$ and $q$ would be imaginary, though. Usually, in these problems, any given parameters are real (or sometimes complex, but usually not if one of them is multiplied by $i$).

Edit (because I can't comment yet due to low reputation): As per your answer, it seems to me that you miswrote something in your original post. Given your equation relating $p$ and $q$, I imagine you meant to say that $p,q\in\mathbb{R}\backslash\{0\}$. If this is the case, then, indeed, $3p=7q$ if $z\in\mathbb{I}$. Also, unless $p$ and $q$ must also be integers, the lower bound for $|z|^2$ is $0$, since I can make $q$ (and therefore $p$) as small as I like without ever reaching $q=0$. I think you should mention that $p,q\in\mathbb{Z}\backslash\{0\}$ or, if they must also be positive, $p,q\in\mathbb{N}$.


I got it now.Thanks for your answers.I can do it like $z = (3p – 7q) + i(3q + 7p)$ .For purely imaginary $3p = 7q$

$p = 7 or q = 3$ (for least value)

Hence $|z| = |3 + 7i| |p + iq|.|z|^2 = 58(p^2 + q^2 ) = 58[72 + 9] = 58^2=3364$


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