A problem on complex no. Q.If $z$ and $w$ be two complex no, such that $|z|=|w|=1$ and$|z+iw|=|z-iw|=2$.Then find z
I tried to break $z$ and $w$ into $a+ib$ form separately and carry on linear equation by substituting the values.But,It leads to me nowhere.Any hints?

Let z = a + bi
Let w = c + di
iw = -d + ci
z + iw = (a-d) + i(b+c)
z - iw = (a+d) +i(b-c)
|z| = |w| = 1  => $a^2 + b^2 = c^2 + d^2 = 1$
|z + iw| =2 => $(a - d)^2 + (b + c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 - 2ad + 2bc =4$ so $2 + 2bc - 2ad =4$ so $bc - ad = 1$
Likewise |z - iw| =2 => $(a + d)^2 + (b - c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 + 2ad - 2bc =4$ so $2 + 2bc - 2ad =4$ so $ad - bc = 1$
So we have $bc - ad = 1$ and  $ad - bc = 1$ which is impossible so, either it's late and I made a stupid error, or ... there's no solution.
