# If z and w are two complex numbers such that |Z| ≤ 1, |W|≤ 1 and |Z + iw| Then, z is equal to (1) 1 or i (2) i or - i (3) 1 or - 1 (4) i or -1

Solution:

Given |Z| ≤ 1and |W|≤ 1

|Z + iw| = $$\left | z-i\bar{w} \right |=2$$

Let Z = x1+iy

w = x2+iy2

Hence x12+y12 ≤ 1

x22+y22 ≤ 1

So |Z+iw| = |x1+y1+i(x2+iy2)| = 2

(x1-y1)2+(y2+x2)2 = 4 ….(1)

Also |Z+iw| = |x1+y1+i(x2+iy2)| = 2

(x1-y2)2+(y1-x2)2 = 4 ….(2)

(2) -(1)

=> (y1-x2)2-(y1+x2)2 = 0

y12 +x22-2y1x2-y12-x22-2y1x2 = 0

y1x2 = 0

y1 = 0 => x12 ≤1

Therefore -1≤x≤1

So Z = 1+i0 or -1+i0

Hence option (3) is the answer.