Thread: The Fundamental Theorem Of Algebra

The Fundamental Theorem Of Algebra
Any polynomial of degree with complex coefficient has (complex) zeros, provided we count the zeros with their multiplicities.
In particular, from The Fundamental Theorem f Algebra follows that any quadratic polynomial with complex coefficients has two complex zeros. This particular case of the FTA is not difficult to prove by employing the Quadratic Formula, provided we can prove that there are two values of the square root of any (nonzero) complex number. Your Mini-Project is particular case of this statement which is true generally.

Problem Find the square roots of -5+12i, i.e. solve the equation z^2= -5+12i by setting the system of equations for the real part x and the imaginary part Y of the solution z = x+yi. Remember that X and Yare real numbers.

Originally Posted by NeoSonata

The Fundamental Theorem Of Algebra
Any polynomial of degree with complex coefficient has (complex) zeros, provided we count the zeros with their multiplicities.

"Any polynomial of degree n with complex coefficient has n (complex) zeros, provided we count the zeros with their multiplicities."

In particular, from The Fundamental Theorem f Algebra follows that any quadratic polynomial with complex coefficients has two complex zeros. This particular case of the FTA is not difficult to prove by employing the Quadratic Formula, provided we can prove that there are two values of the square root of any (nonzero) complex number. Your Mini-Project is particular case of this statement which is true generally.

Problem Find the square roots of -5+12i, i.e. solve the equation z^2= -5+12i by setting the system of equations for the real part x and the imaginary part Y of the solution z = x+yi. Remember that X and Yare real numbers.

Excuse me? This is,as it says, your mini- project- you are to do it yourself. If z= x+ iy, what is z^2? Set that equal to -5+ 2i and equate real and imaginary parts on each side.

Write z as x + yi and square it, getting x^2 - y^2 + 2xyi, and equate this to -5 + 12i, thus x^2-y^2 = -5 and 2xy = 12.

z= x + yi

z^2 = (x + yi)^2

z^2 = x^2-y^2+ 2xyi

-5+12i = x^2 - y^2 + 2xyi

Can you explain why did you separate it into x^2-y^2 = -5 and 2xy= 12?

Is it because they ask you to set the system of equations for real part X and imaginary part Y?
Since they are asking you to square the problem, those 2 separate equations gives the + - answer?

Originally Posted by NeoSonata

z= x + yi

z^2 = (x + yi)^2

z^2 = x^2-y^2+ 2xyi

-5+12i = x^2 - y^2 + 2xyi

Can you explain why did you separate it into x^2-y^2 = -5 and 2xy= 12?

Is it because they ask you to set the system of equations for real part X and imaginary part Y?
Since they are asking you to square the problem, those 2 separate equations gives the + - answer?

That is what the problem tells you to do: "setting the system of equations for the real part x and the imaginary part Y of the solution z = x+yi."

Since they are asking you to square the problem, those 2 separate equations gives the + - answer?

Find every solution for x and y. If you have any doubt about whether a given result is indeed a solution, plug these values back into the original equation to check.

Okay thank you Ragnar, I don't doubt whether your answer is correct or not. I just have trouble understanding how you formed 2 equation from 1.

No worries, I understand.

Presumably you have no difficulty arriving at the equation x^2 - y^2 + 2xyi = -5 + 12i. Since the x^2 +y^2 has no imaginary element (remember x and y are pure real numbers), it must provide the real part of -5 + 12i and thus x^2 + y^2 = -5. Likewise 2xyi must be the imaginary part, again because x and y are pure real numbers, and so is equal to the imaginary part of -5 + 12i, that is to say it is equal to 12. That is how we get the two equations equate the real part of x^2 - y^2 + 2xyi with the real part of -5 + 12i obtaining x^2 - y^2 = -5, and the same for the imaginary part to get 2xy = 12.

yes I finally understand it. Thanks again!