Let z and w be two complex numbers such that |z|, le 1, |w|, le 1 and |z + iw|, = ,|z

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4-1.Complex numbers

hard

Let $z$ and $w$ be two complex numbers such that $|z|\, \le 1,$ $|w|\, \le 1$and $|z + iw|\, = \,|z - i\overline w | = 2$. Then $z$ is equal to

$1$ or $i$

$i$ or $ - i$

$1$ or $-1$

$i$ or $ -1$

(IIT-1995)

Solution

(c)Let $z = a + ib,|z| \le 1$$⇒$ ${a^2} + {b^2} \le 1$
and $w = c + id,|w|\, \le 1$$⇒$ ${c^2} + {d^2} \le 1$
$|z + iw|\, = \,|a + ib + i(c + id)| = 2$
$⇒$ ${(a – d)^2} + {(b + c)^2} = 4$ ……$(i)$
$|z – i\overline w |\, = \,|a + ib – i(c – id)|$
$⇒$ ${(a – d)^2} + {(b – c)^2} = 4$……$(ii)$
From $(i) $ and $ (ii)$, we get $bc = 0$
$⇒$ Either $b = 0$or $c = 0$
If $b = 0$, then ${a^2} \le 1$.

Then, only possibility is $a = 1$or $ – 1$.

Standard 11

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medium

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