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Question

If 5z17z2 is purely imaginary, then 2z1+3z22z13z2is

A
57
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B
75
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C
1
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D
35
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Solution

The correct option is C 1
57z1z2 is purely imaginary z1z2 is purely imaginary

z1z2+¯z1¯z2=0z1¯z2+z2¯z1=0(1)

Now,
2z1+3z22z13z22=(2z1+3z2)×(2¯z1+3¯z2)(2z13z2)×(2¯z13¯z2){as|z|2=z.¯z}

=4|z1|2+6z1¯z2+6z2¯z1+9|z22|4|z1|26z1¯z26z2¯z1+9|z2|2

=4|z1|2+9|z2|2+6(z1¯z2+z2¯z1)4|z1|2+9|z2|26(z2¯z1+z1¯z2)

But from (1) ,z1¯z2+z2¯z1=0

=4|z1|2+9|z2|24|z1|2+9|z2|2=1

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