Z=(1-2i)(2-i)

Solution for Z=(1-2i)(2-i) equation:

=(1-2Z)(2-Z)

We move all terms to the left:

-((1-2Z)(2-Z))=0

We add all the numbers together, and all the variables

-((-2Z+1)(-1Z+2))=0

We multiply parentheses ..

-((+2Z^2-4Z-1Z+2))=0


We calculate terms in parentheses: -((+2Z^2-4Z-1Z+2)), so:

(+2Z^2-4Z-1Z+2)

We get rid of parentheses

2Z^2-4Z-1Z+2

We add all the numbers together, and all the variables

2Z^2-5Z+2

Back to the equation:

-(2Z^2-5Z+2)


We get rid of parentheses

-2Z^2+5Z-2=0

a = -2; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·(-2)·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*-2}=\frac{-8}{-4}
=+2
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*-2}=\frac{-2}{-4}
=1/2
$