## Trigonometry (11th Edition) Clone

$z_1^2 - 1$ = $(a^2 - b^2 - 1) + 2abi$ $z_2^2 - 1$ = $(a^2 - b^2 - 1) - 2abi$
For $z_1 = a + bi$, $z_1^2 - 1$ = $(a + bi)^2 - 1$ = $a^2 + 2abi + (bi)^2 - 1$ = $(a^2 - b^2 - 1) + 2abi$ (since $i^2 = -1$) For $z_2 = a - bi$, $z_2^2 - 1$ = $(a - bi)^2 - 1$ = $a^2 - 2abi + (bi)^2 - 1$ = $(a^2 - b^2 - 1) - 2abi$ (since $i^2 = -1$)