How Do You Correctly Multiply Complex Numbers to Verify Roots?

[Square1]

I have solved the roots of a quadratic equation and want to "test" them by putting them back in for x. I am having a problem with the x^2 term. Of the two roots, I'm only trying so far the positive square root case. I am trying to avoid writing all my work out since that would be hell and I also think would not be easy to read.

To evaluate x^2 where x = -2/3 + i√19, I first try the "from the ground up" method of just distributing, and eventually make a substitution of -1 for i^2. After making that substitution, I end up subtracting the term that had the i^2, from (9/4). After simplifying, I have (-5/2) + (-3i√19)/2

The second way I try to evaluate x^2 term is by using the multiplication definition of complex numbers (a + bi)(c + di) = (ac - bd) + (bc + ad)i . I find that (ac - bd) results in (9/4) - (19i^2)/4 and when I change i^2 to -1, now the terms are being added! Then when I add the (bc + ad)i part, I finally end up with (14/2) + (-3i√19)/2 .

Could someone point out the mistake here then? :S

 

[Square1]

I am new to complex numbers so don't assume anything with me. I read somewhere on wiki that it is sometimes relevant for a +bi when b is negative to keep b itself negative as supposed to write as a - bi ? I don't really remember about it much though, but just throwing it out there.

 

[Mark44]

I have solved the roots of a quadratic equation and want to "test" them by putting them back in for x. I am having a problem with the x^2 term. Of the two roots, I'm only trying so far the positive square root case. I am trying to avoid writing all my work out since that would be hell and I also think would not be easy to read.

If we can't see what you did, how can we provide help?

To evaluate x^2 where x = -2/3 + i√19, I first try the "from the ground up" method of just distributing, and eventually make a substitution of -1 for i^2. After making that substitution, I end up subtracting the term that had the i^2, from (9/4). After simplifying, I have (-5/2) + (-3i√19)/2

This is incorrect. Since you didn't show your work, I have no way to point out where you went wrong.

The second way I try to evaluate x^2 term is by using the multiplication definition of complex numbers (a + bi)(c + di) = (ac - bd) + (bc + ad)i . I find that (ac - bd) results in (9/4) - (19i^2)/4 and when I change i^2 to -1, now the terms are being added!

You have mistake here, as well, in both calculations.

a = c = -2/3, and b = d = √19. Note that b and d are the coefficients of i. They do not include i.

ac - bd = 4/9 - 19

Then when I add the (bc + ad)i part, I finally end up with (14/2) + (-3i√19)/2 .

Could someone point out the mistake here then? :S

What do you get for bc + ad? That is the coefficient of i in the product.

 

[Square1]

Ok i made some big mistakes in writing this firstly. Yikes. Need to start again...

The roots of x^2 + 3x + 7 are x = -3/2 + (i√19)/2 and x = -3/2 - (i√19)/2

x^2 when x = -3/2 + (i√19)/2
= (-3 + i√19)/(2) x (-3 + i√19)/(2)
= (9/4 - (3i√19)/4 - (3i√19)/4 - 19/4) switching i^2 with -1
=(-10/4 - (6i√19)/4)
=(-5/2 - (3i√19)/2)Now plugging in terms with the definition...
( (9/4 - (19i^2)/4) + ( (-3i√19)/4 - (3i√19)/4 ) )
= ( (9 + 19)/4 + (-6i√19)/4 )
= 28/4 - (6i√19)/4
=14/2 - (3i√19)/4

 

[SteamKing]

Try this: (a+b)^2 = a^2 + 2ab + b^2
((-3/2) + i*SQRT(19)/2)^2 = (-3/2)^2 + 2(3/2)(i*SQRT(19)/2) + (19/4)*i^2)
= 9/4 + (-1)*(19/4) + (3/2)(i*SQRT(19)/2
= (9-19)/4 + (3/4)*SQRT(19)*i
= -5/2 + 3i*SQRT(19)/4

You have to be careful handling - signs and i^2 in the same expression, or you might inadvertently make a mistake.

 

[Square1]

OK i think I got it now.
I had the right idea with my distributive method and it brings me to right answer. When it comes to just plugging in the needed values "using the definition", it boiled down to what mark mentioned; that being to exclude i from term b and d.

Thanks