Q.
If Im(z) of z−1eiθ+eiθz−1 is zero, then z lies on
1934
220
Complex Numbers and Quadratic Equations
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Solution:
Let t=z−1eiθ
Given Im(z) of (t+t1)=0 ∴(t+t1)−(t+t1)=0 (∵z=x+iy=x,∴zˉ=x⇒z−zˉ=0) ⇒(t−tˉ)+(t1−t1)=0 ⇒(t−tˉ)+ttˉtˉ−t=0 ⇒(t−tˉ)(1−∣t∣21)=0 ∴1−∣t∣21=0 or t=tˉ ⇒∣t∣2=1⇒∣∣z−1eiθ∣∣2=1 ⇒∣z−1∣2=1 (∵∣∣eiθ∣∣=∣cosθ+isinθ∣=1) ⇒∣z−1∣=1 or (x−1)2+y2=1
which represent a circle of unit radius