(1+2j)*(1-2j)=(5+0j)

Solution for (1+2j)*(1-2j)=(5+0j) equation:

(1+2j)(1-2j)=(5+0j)

We move all terms to the left:

(1+2j)(1-2j)-((5+0j))=0

We add all the numbers together, and all the variables

(2j+1)(-2j+1)-((j+5))=0

We multiply parentheses ..

(-4j^2+2j-2j+1)-((j+5))=0


We calculate terms in parentheses: -((j+5)), so:

(j+5)

We get rid of parentheses

j+5

Back to the equation:

-(j+5)


We get rid of parentheses

-4j^2+2j-2j-j+1-5=0

We add all the numbers together, and all the variables

-4j^2-1j-4=0

a = -4; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·(-4)·(-4)
Δ = -63
Delta is less than zero, so there is no solution for the equation