Question
If $${\text{amp}}\frac{{z - 2}}{{2z + 3i}} = 0$$ and $${z_0} = 3 + 4i$$ them
A.
$${z_0}\overline z + {\overline z _0}z = 12$$
B.
$${z_0}z + {\overline z _0}\overline z = 12$$
C.
$${z_0}\overline z + {\overline z _0}z = 0$$
D.
None of these
Answer :
$${z_0}z + {\overline z _0}\overline z = 12$$
Solution :
$$\eqalign{
& \frac{{z - 2}}{{2z + 3i}} = \frac{{\left( {x - 2} \right) + iy}}{{2x + i\left( {2y + 3} \right)}} \cr
& \therefore \,\,{\text{amp}}\frac{{z - 2}}{{2z + 3i}} = {\text{amp}}\left\{ {\left( {x - 2} \right) + iy} \right\} - {\text{amp}}\left\{ {2x + i\left( {2y + 3} \right)} \right\} \cr
& {\text{or, }}0 = {\tan ^{ - 1}}\frac{y}{{x - 2}} - {\tan ^{ - 1}}\frac{{2y + 3}}{{2x}} \cr
& \Rightarrow \,\,\frac{y}{{x - 2}} = \frac{{2y + 3}}{{2x}} \cr
& \Rightarrow \,\,3x - 4y = 6. \cr
& {\text{Now, }}{z_0}z = \left( {3 + 4i} \right)\left( {x + iy} \right) = \left( {3x - 4y} \right) + i\left( {4x + 3y} \right) \cr
& {z_0}\overline z = \left( {3 + 4i} \right)\left( {x - iy} \right) = \left( {3x + 4y} \right) + i\left( {4x - 3y} \right) \cr
& {\overline z _0}z = \left( {3 - 4i} \right)\left( {x + iy} \right) = \left( {3x + 4y} \right) + i\left( { - 4x + 3y} \right) \cr
& {\overline z _0}\overline z = \left( {3 - 4i} \right)\left( {x - iy} \right) = \left( {3x - 4y} \right) + i\left( { - 4x - 3y} \right). \cr} $$
Check the correct answer.