Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.2 Trigonometric (Polar) Form of Complex Numbers - 8.2 Exercises - Page 371: 68b

$z_1^2 - 1$ = $(a^2 - b^2 - 1) + 2abi$ $z_2^2 - 1$ = $(a^2 - b^2 - 1) - 2abi$

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 $i^2 = -1$)