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# If z = (3 + 7i) (a + ib) where $\mathrm{a},\mathrm{b}\in \mathrm{Z}-\left\{0\right\}$, is purely imaginary, then the minimum value of |z|2 is

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58
d
65

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detailed solution

Correct option is C

z=(3a−7b)+i(3b+7a)For z to be purely imaginary we have3a=7b⇒a=7 or b=3            (for least value of |z|)⇒|z|=9+49=58

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Similar Questions

Let  and $\mathrm{i}=\sqrt{-1}$

then  $\mathrm{r}=\sqrt{\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}=|\mathrm{z}|$

and  $\mathrm{\theta }={\mathrm{tan}}^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)=\mathrm{arg}\left(\mathrm{z}\right)$

now, $|\mathrm{z}{|}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{ib}\right)\left(\mathrm{a}-\mathrm{ib}\right)=\mathrm{z}\overline{\mathrm{z}}$

and  $\left|{\mathrm{z}}_{1}{\mathrm{z}}_{2}{\mathrm{z}}_{3}\dots .{\mathrm{z}}_{\mathrm{n}}\right|=\left|{\mathrm{z}}_{1}\right|\left|{\mathrm{z}}_{2}\right|\left|{\mathrm{z}}_{3}\right|\dots \left|{\mathrm{z}}_{\mathrm{n}}\right|$

if  then f(z) is called unimodular. In this case
f(z) can always be expressed as $\mathrm{f}\left(\mathrm{z}\right)={\mathrm{e}}^{\mathrm{i\alpha }},\mathrm{\alpha }\in \mathrm{R}$

Also, ${\mathrm{e}}^{\mathrm{i\alpha }}+{\mathrm{e}}^{\mathrm{i\beta }}={\mathrm{e}}^{\mathrm{i}\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}2\mathrm{cos}\left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$  and

${\mathrm{e}}^{\mathrm{i\alpha }}-{\mathrm{e}}^{\mathrm{i\beta }}={\mathrm{e}}^{\mathrm{i}\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}2\mathrm{isin}\left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$ where  $\mathrm{\alpha },\mathrm{\beta }\in \mathrm{R}$