If {Z_1} ne 0 and {Z_2} are two complex numbe | vedclass

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(D) Let $\frac{{{Z_2}}}{{{Z_1}}} = ki$,where $k \in \mathbb{R}$ and $k 
e 0$.Then,$\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} ight| = \le

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(D) Let $\frac{{{Z_2}}}{{{Z_1}}} = ki$,where $k \in \mathbb{R}$ and $k 
e 0$.Then,$\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} ight| = \left| {\frac{{{Z_1}(2 + 3\frac{{{Z_2}}}{{{Z_1}}})}}{{{Z_1}(2 - 3\frac{{{Z_2}}}{{{Z_1}}})}}} ight| = \left| {\frac{{2 + 3ki}}{{2 - 3ki}}} ight|$.Since the modulus of a complex number $z$ is equal to the modulus of its conjugate $\bar{z}$,and for any complex number $w$,$|w| = |\bar{w}|$,we have:$\left| {\frac{{2 + 3ki}}{{2 - 3ki}}} ight| = \frac{{|2 + 3ki|}}{{|2 - 3ki|}} = \frac{{\sqrt {{2^2} + {{(3k)}^2}} }}{{\sqrt {{2^2} + {{( - 3k)}^2}} }} = \frac{{\sqrt {4 + 9{k^2}} }}{{\sqrt {4 + 9{k^2}} }} = 1$.

If ${Z_1} \ne 0$ and ${Z_2}$ are two complex numbers such that $\frac{{{Z_2}}}{{{Z_1}}}$ is a purely imaginary number,then $\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} \right|$ is equal to

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