If {Z_1} ne 0 and Z_2 be two comple | vedclass

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Question

Let ${	ext{ }}{z_1} = 1 + i{	ext{ }}$ and ${z_2} = 1 - i$

$\frac{{{z_2}}}{{{z_1}}} = \frac{{1 - i}}{{1 + i}} $ $= \frac{{(1 - i)(1 - i)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

Let ${	ext{ }}{z_1} = 1 + i{	ext{ }}$ and ${z_2} = 1 - i$

$\frac{{{z_2}}}{{{z_1}}} = \frac{{1 - i}}{{1 + i}} $ $= \frac{{(1 - i)(1 - i)}}{{(1 + i)(1 - i)}}\, = \, - \,i$

$\frac{{2{z_1} + 3{z_2}}}{{2{z_1} - 3{z_2}}} = $ $\frac{{2 + 3\left( {\frac{{{z_2}}}{{{z_1}}}} ight)}}{{2 - 3\left( {\frac{{{z_2}}}{{{z_1}}}} ight)}}$ $ = \frac{{2 - 3i}}{{2 + 3i}}$

$\left| {\frac{{2{z_1} + 3{z_2}}}{{2{z_1} - 3{z_2}}}} ight| = \,\left| {\frac{{2 - 3i}}{{2 + 3i}}} ight|$ $ = \,\left| {\frac{{2 - 3i}}{{2 + 3i}}} ight|$

$\left[ {\because \left| {\frac{{{z_1}}}{{{z_2}}}} ight| = \frac{{\left| {{z_1}} ight|}}{{\left| {{z_2}} ight|}}} ight]$

$ = \frac{{\sqrt {4 + 9} }}{{\sqrt {4 + 9} }} = 1$

If ${Z_1} \ne 0$ and $Z_2$ be 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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