Complex NumbersVery Hard
Question
Let z and w be two complex number such that |z| ≤ 1, |w|1 and |z + iw| = |z - 
| = 2, then z equals
| = 2, then z equals Options
A.1 or i
B.i or - i
C.1 or - 1
D.i or - 1
Solution
Given, |z + iw| = |z - | = 2
⇒ |z - (-iw)| = |z - (i) = 2
⇒ |z - (-iw)| = |z - (-i)|
∴ lies on the perpendicular of the line joining - iw and - i. Since - i is the mirror image of - iw in the x - axis, the locuos of z is the x - axis.
Let z = x + iy and y = 0.
Now, |z| |z| ≤ 1 ⇒ x2 + 02 ≤ 1 ⇒ - 1 ≤ x ≤ 1.
∴ z may take values given in (c).
| = 2⇒ |z - (-iw)| = |z - (i
) = 2⇒ |z - (-iw)| = |z - (-i
)| ∴ lies on the perpendicular of the line joining - iw and - i
. Since - i is the mirror image of - iw in the x - axis, the locuos of z is the x - axis.Let z = x + iy and y = 0.
Now, |z| |z| ≤ 1 ⇒ x2 + 02 ≤ 1 ⇒ - 1 ≤ x ≤ 1.
∴ z may take values given in (c).
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