Help me factorize a quadratic equation in a complex variable

[aristotle_sind]

The equation is: (appears while solving a trigonometric integral using residue theorem)

2Z²+iZ²-6Z+2-i
=(2+i)Z²-6Z+(2-i)

The roots are:
Z₁=(2-i) and Z₂=(2-i)/5

I can't write the equation in factored form.
If I simply write it like this:
{z-(2-i)}{5z-(2-i)}
It doesn't give the same equation back i.e.
{z-(2-i)}{5z-(2-i)}=5Z²-12Z+6iZ+3-4i
which is not the equation that I started with.

Can anyone please write the equation in factored form...

 

[BvU]

Did you notice PF has a template ?

In the first place, I don't see an equation. But I suppose you want to solve 2Z²+iZ²-6Z+2-i = 0 ?

You claim a root is z = 2-i. How do you know ?

Let me give you a hint, anyway: what is 5/(2+i) ?

 

[ehild]

In the first place, I don't see an equation. But I suppose you want to solve 2Z²+iZ²-6Z+2-i = 0 ?

You claim a root is z = 2-i. How do you know ?

Let me give you a hint, anyway: what is 5/(2+i) ?

The roots 2-i and (2-i)/5 are correct. They are the same as 5/(2+i) and 1/(2+i).

ehild

 

[ehild]

The equation is: (appears while solving a trigonometric integral using residue theorem)

2Z²+iZ²-6Z+2-i
=(2+i)Z²-6Z+(2-i)

The roots are:
Z₁=(2-i) and Z₂=(2-i)/5

I can't write the equation in factored form.
If I simply write it like this:
{z-(2-i)}{5z-(2-i)}
It doesn't give the same equation back i.e.


Can anyone please write the equation in factored form...

az²+bz+c=a(z-z1)(z-z2)

You forgot the coefficient of z².

ehild

 

[BvU]

Dear ehild, what a splurge of posts. Why don't we let poor old Ari contemplate a bit ?

 

[ehild]

Dear ehild, what a splurge of posts. Why don't we let poor old Ari contemplate a bit ?

The OP solved the quadratic equation corresponding to the original formula correctly. Why did you try to confuse him?

ehild

 

[BvU]

Oh boy, that wasn't the intention. The intention was to let Ari see that a simple multiplication of the coefficients with (2+i)/5 would let the one eq go over into the other.

Maybe it would even have worked! Or does Ari have the feeling I tried to confuse him/her ?

 

[ehild]

You asked the OP how did he know that 2-i was a solution. Was not it confusing?

ehild

 

[aristotle_sind]

Thank you for the replies...

Did you notice PF has a template ?

In the first place, I don't see an equation. But I suppose you want to solve 2Z²+iZ²-6Z+2-i = 0 ?

You claim a root is z = 2-i. How do you know ?

Let me give you a hint, anyway: what is 5/(2+i) ?

Well .. my bad .. that certainly wasn't a good way to ask for help and my equation looked weird. I've bookmarked the guidelines thread .. will refer to it before making a new thread in future. Thank you for that...

And I did get a root that was 5/(2+i) but just wanted to get rid of i in the denominator.

az²+bz+c=a(z-z1)(z-z2)

You forgot the coefficient of z².

ehild

Thank you .. Didn't even realize I was making this mistake. Had completely forgotten that you add the coefficient of Z² while factorizing a quadratic.