Q.
If z1 and z2 both satisfy z+zˉ=2∣z−1∣ , arg (z1−z2)=4π , then find Im (z1+z2) .
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219
Complex Numbers and Quadratic Equations
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Solution:
Let z=x+iy , z1=x1+iy1
and z2=x2+iy2, z+zˉ=2∣z−1∣ ⇒(x+iy)+(x−iy) =2∣x−1+iy∣ ⇒2x=1+y2...(i)
Since z1 and z2 both satisfy (i), we have 2x1=1+y12 and 2x2=1+y22 ⇒2(x1−x2)=(y1+y2)(y1−y2) ⇒2=(y1+y2)(x1−x2y1−y2)…(ii)
Again z1−z2=(x1−x2)+i(y1−y2)
Therefore, tanθ=x1−x2y1−y2 ,
where θ=arg(z1−z2) ⇒tan4π=x1−x2y1−y2 (since θ4π)
i.e., 1=x1−x2y1−y2
From (ii) , we get 2=y1+y2 ,
i.e., Im (z1+z2)=2