MF26

We have that $(2-3i)^2 = -5-12i$
$(z-i+1)^2 = 5+12i$
$(z-i+1)^2 = -(-5-12i)$
$(z-i+1)^2 = -(2-3i)^2$
$(z-i+1)^2 = i^2 (2-3i)^2$
$(z-i+1)^2 = [i(2-3i)]^2$
$(z-i+1)^2 = (2i+3)^2$
$(z-i+1)^2 = (2i+3)^2$
$(z-i+1) = \pm(2i+3)$
$z = \pm(2i+3) - 1 + i$
$z = 2i+3 - 1 + i$ or $z = -(2i+3) - 1 + i$
$z = 3i+2$ or $z = -i+2$

This is a worksheet from class.
There is a slight typo, we need to show that $x^2 (x+2)^2 - 1 = (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})$
So for all proof/ show questions, it is important that we can identify whether we should prove from LHS to RHS or RHS to LHS.
For this question, we observe that if we start from LHS, we are trying to expand our expression. But if we start from RHS, we are trying to simplify our expression. It will be generally easier to simplify.

$RHS$
$= (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})$
$= (x+1)^2[(x+1)^2 - (\sqrt{2})^2]$
$= (x+1)^2[(x+1)^2 - 2]$
$= (x+1)^2(x^2 + 2x + 1 - 2]$
$= (x^2 + 2x + 1)(x^2 + 2x -1)$; we can also expand it fully
$= [x(x+2)+1][x(x+2)-1]$
$= x^2 (x+2)^2 -1$

Here’s a much faster but less obvious method!
$LHS$
$= x^2 (x+2)^2 -1$
$= [x(x+2)+1][x(x+2)-1]$
$= (x^2 + 2x + 1)(x^2 + 2x -1)$; using quadratic formula
$= (x+1)^2 (x - \frac{-2-\sqrt{8}}{2})x - \frac{-2+\sqrt{8}}{2})$
$= (x+)^2(x+1+1+\sqrt{2})(x+1-\sqrt{2})$

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