Question

If $z=(3+7i)(p+iq)$ where $p,q\in 1-\left\{0\right\}$, is purely imaginary, then minimum value of ${\left|z\right|}^2$ is

• A. $0$
• B. $58$
• C. $\frac{3364}{9}$
• D. $3364$

Answer 1

The correct option is D $3364$

Given:$z=(3+7i)(p+iq)$

$=3(p+iq)+7i(p+iq)$

$=3p+3iq+7ip+7i^{2}q$

$=3p+3iq+7ip-7q$ since $i^2=-1$

$z=(3p-7q)+i(3q+7p)$

where $x=3p-7q$ and $y=3q+7p$

$=(3p-7q)+i(3q+7p)$ is imaginary.

$\Rightarrow 3p-7q=0$

$\Rightarrow p=\frac{7q}{3}$

${\left|z\right|}^2=x^2+y^2$

${\left|z\right|}^2=0+{(3q+7p)}^2$

${\left|z\right|}^2={(3q+7\times \frac{7q}{3})}^2$ where $p=\frac{7q}{3}$

${\left|z\right|}^2={(3q+\frac{49q}{3})}^2$

${\left|z\right|}^2={\left(\frac{9q+49q}{3}\right)}^2$

${\left|z\right|}^2={\left(\frac{58q}{3}\right)}^2$

${\left|z\right|}^2=\frac{3364q^2}{9}$ for $q\in I-\left\{0\right\}$

Minimum value $=\frac{3364\times 3^2}{9}$ for $q=3$

$\therefore$Minimum value $=3364$