Let z and w be two complex numbers such that | vedclass

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(C) Given $|z| \le 1$ and $|w| \le 1$.We are given $|z + iw| = 2$ and $|z - i\overline{w}| = 2$.Let $z = a + ib$ and $w = c + id$. Then $|z|^2 = a^2 +

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$1$ or $-1$

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(C) Given $|z| \le 1$ and $|w| \le 1$.We are given $|z + iw| = 2$ and $|z - i\overline{w}| = 2$.Let $z = a + ib$ and $w = c + id$. Then $|z|^2 = a^2 + b^2 \le 1$ and $|w|^2 = c^2 + d^2 \le 1$.$|z + iw| = |(a + ib) + i(c + id)| = |(a - d) + i(b + c)| = 2$.Squaring both sides: $(a - d)^2 + (b + c)^2 = 4$ $(i)$.$|z - i\overline{w}| = |(a + ib) - i(c - id)| = |(a - d) + i(b - c)| = 2$.Squaring both sides: $(a - d)^2 + (b - c)^2 = 4$ $(ii)$.Subtracting $(ii)$ from $(i)$,we get $(b + c)^2 - (b - c)^2 = 0$,which simplifies to $4bc = 0$,so $bc = 0$.If $b = 0$,then $(a - d)^2 + c^2 = 4$. Since $a^2 \le 1$ and $c^2 + d^2 \le 1$,the maximum value of $(a - d)^2 + c^2$ is $(1 - (-1))^2 + 1 = 5$ (not helpful). However,for $(a - d)^2 + c^2 = 4$ with $a^2 \le 1$ and $c^2 + d^2 \le 1$,the only way to reach $4$ is if $a = 1, d = -1, c = 0$ or $a = -1, d = 1, c = 0$.In both cases,$z = a + i(0) = \pm 1$.

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

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