If P and Q are represented by the complex numbers z1 and z2 respectively, such that ∣∣∣1z2+1z1∣∣∣=∣∣∣1z2−1z1∣∣∣, then
A
△OPQ (where O is the origin) is equilateral
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B
△OPQ is right angled
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C
The circumcentre of △OPQ is 12(z1+z2)
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D
The circumcentre of △OPQ is 13(z1+z2)
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Solution
The correct option is C The circumcentre of △OPQ is 12(z1+z2) We have, ∣∣∣1z2+1z1∣∣∣=∣∣∣1z2−1z1∣∣∣⇒|z1+z2|=|z1−z2|
Squaring both sides, we get ⇒|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2)=|z1|2+|z2|2−2Re(z1¯¯¯¯¯z2)⇒|z1|2+|z2|2+2(z1¯¯¯¯¯z2+¯¯¯¯¯z1z2)=|z1|2+|z2|2−2(z1¯¯¯¯¯z2+¯¯¯¯¯z1z2)⇒4(z1¯¯¯¯¯z2+¯¯¯¯¯z1z2)=0⇒z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=0⇒arg(z1z2)=±π2=arg(z1−0z2−0)