(1+3i)(1-3i)=0

Solution for (1+3i)(1-3i)=0 equation:

(1+3i)(1-3i)=0

We add all the numbers together, and all the variables

(3i+1)(-3i+1)=0

We multiply parentheses ..

(-9i^2+3i-3i+1)=0

We get rid of parentheses

-9i^2+3i-3i+1=0

We add all the numbers together, and all the variables

-9i^2+1=0

a = -9; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-9)·1
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-9}=\frac{-6}{-18}
=1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-9}=\frac{6}{-18}
=-1/3
$