7
Question
Show that the equation of a straight line in the Argand plane can put in the form z¯¯b+b¯¯¯z=c, where b is a non zero complex constant and c is real.
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Solution
Let z1,z2 be two given points A, B on the argand plane.
Let P(z) be any point on the line AB.
∵ A, P, B are collinear, ∴arg(z−z1z1−z2)=0 or ±π
so that z−z1z1−z2 is purely real.
i.e., Im(z−z1z1−z2)=0
⇒z−z1z1−z2−¯¯¯z−¯¯¯¯¯z1¯¯¯¯¯z1−¯¯¯¯¯z2=0
⇒z(¯¯¯¯¯z1−¯¯¯¯¯z2)−¯¯¯¯¯z1(z1−z2)+(z1¯¯¯¯¯z2−¯¯¯¯¯z1z2)=0 ...........(1)
Now since z1¯¯¯¯¯z2 is conjugate to ¯¯¯¯¯z1z2, the number
z1¯¯¯¯¯z2−¯¯¯¯¯z1z2 is purely imaginary
Let z1¯¯¯¯¯z2−¯¯¯¯¯z1z2=ic ............ (2)
then from (1) and (2)
z(¯¯¯¯¯z1−¯¯¯¯¯z2)−¯¯¯z(z1−z2)+ic=0
Multiplying by i
⇒zi(¯¯¯¯¯z1−¯¯¯¯¯z2)−i¯¯¯z(z1−z2)−c=0 ........... (3)
If i(z2−z1)=b then ¯¯b=¯i(¯¯¯¯¯z2−¯¯¯¯¯z1)
=−i(¯¯¯¯¯z2−¯¯¯¯¯z1)
=i(¯¯¯¯¯z1−¯¯¯¯¯z2)
then equation (3) reduced in the form
z¯¯b+¯¯¯zb=c
Ans: 1
Let P(z) be any point on the line AB.
∵ A, P, B are collinear, ∴arg(z−z1z1−z2)=0 or ±π
so that z−z1z1−z2 is purely real.
i.e., Im(z−z1z1−z2)=0
⇒z−z1z1−z2−¯¯¯z−¯¯¯¯¯z1¯¯¯¯¯z1−¯¯¯¯¯z2=0
⇒z(¯¯¯¯¯z1−¯¯¯¯¯z2)−¯¯¯¯¯z1(z1−z2)+(z1¯¯¯¯¯z2−¯¯¯¯¯z1z2)=0 ...........(1)
Now since z1¯¯¯¯¯z2 is conjugate to ¯¯¯¯¯z1z2, the number
z1¯¯¯¯¯z2−¯¯¯¯¯z1z2 is purely imaginary
Let z1¯¯¯¯¯z2−¯¯¯¯¯z1z2=ic ............ (2)
then from (1) and (2)
z(¯¯¯¯¯z1−¯¯¯¯¯z2)−¯¯¯z(z1−z2)+ic=0
Multiplying by i
⇒zi(¯¯¯¯¯z1−¯¯¯¯¯z2)−i¯¯¯z(z1−z2)−c=0 ........... (3)
If i(z2−z1)=b then ¯¯b=¯i(¯¯¯¯¯z2−¯¯¯¯¯z1)
=−i(¯¯¯¯¯z2−¯¯¯¯¯z1)
=i(¯¯¯¯¯z1−¯¯¯¯¯z2)
then equation (3) reduced in the form
z¯¯b+¯¯¯zb=c
Ans: 1
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