Complex numbers

\hspace{-16}$Here $\bf{\mathbb{Z}=(3+7i).(p+iq)=(3p-7q)+i.(7p+3q)}$\\\\\\ Now Here $\bf{\mathbb{Z}}$ is Purely Imiginary, Then $\bf{\mathbb{R\text{e}(Z)}=0}$\\\\\\ So $\bf{3p-7q=0\Leftrightarrow p=\frac{7}{3}.q}$ and $\bf{\mathbb{Z}=i.(7p+3q)}$\\\\\\ So $\bf{Min.\mid\mathbb{Z}^2 \mid = Min.\mid\mathbb{Z} \mid^2 = \mid 7p+3q \mid^2}$\\\\\\ Now Here $\bf{p\;,q\in \mathbb{Z}-\left\{0\right\}}$ and $\bf{p=\frac{7}{3}q}$\\\\\\ So We Take $\bf{p_{min}=\pm 3}$ and $\bf{q_{min}=\pm 7}$\\\\\\ Means $\bf{(p,q)=(3,7)}$ or $\bf{(p,q)=(-3,-7)}$\\\\\\ So $\bf{Min.\mid\mathbb{Z} \mid^2 = \mid 7p+3q \mid^2 = \mid 7.7+3.3\mid^2 = \mid -7.7-3.3\mid^2 = 3364}$

z=(3+7i) (p+iq)=(3p-7q)+i(3q+7p)
now it is said z is purely imaginary
so,
3p-7q=0
or p/q=7/3
or p/q + i = 7/3 +i
(p+qi)/q=7+3i/3
from here the p+iq = 7+3i

z= 58i
|z²|= 3364

let me start from your 2nd line
(p+qi)/q=7+3i/3
now can't i say p=14 & q=6 ????
of course u r right in case of minimum.

are yaar tum samjh te kyun nehi q is real and p+iq is imaginary.....
so...
imaginary no. only give imaginary results right thats why
p+iq = 7+3i
p=7
q=3
tum mujhe chat rahe ho??