(1+i)*(1-i)=0

Solution for (1+i)*(1-i)=0 equation:

(1+i)(1-i)=0

We add all the numbers together, and all the variables

(i+1)(-1i+1)=0

We multiply parentheses ..

(-1i^2+i-1i+1)=0

We get rid of parentheses

-1i^2+i-1i+1=0

We add all the numbers together, and all the variables

-1i^2+1=0

a = -1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1)·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*-1}=\frac{-2}{-2}
=1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*-1}=\frac{2}{-2}
=-1
$