Sum of two square and complex quadratic

• r.a.c.
In summary, the first question asks to express 962 as a sum of two squares, which can be solved through brute strength or by finding values of a and b such that a^2 + b^2 = 962. The second question involves finding a complex number a such that the roots of the equation z^2 + az + 1 = 0 have equal absolute values, with one possible solution being a real number. The "auxiliary equation" method is not applicable in this case, and further suggestions are needed to solve it.
r.a.c.
I actually have two questions, both are in the complex plane.

Homework Statement

Q1: Express 962 as a sum of two squares (Hint: 962 = (13)(74)
Q2. Given z,a belong to C (complex). Find a such that the roots of the equation $$z^2 + az + 1 = 0$$ have equal absolute values (or modulus).

Homework Equations

Well this is about the complex plane so for the first one I think most probably its Euler's Identity. As for the second it could be anything.

The Attempt at a Solution

Q1: It is a sum of squares so the best I can think of is that 962 is a modulus of a vector i.e. z belongs to C, |z| = 962. Because z = a + ib so $$962^2 = a^2 + b^2$$. Then I try writing z in polar so we have $$z = |z|e^i^x$$ , where x is the argument of z. But x = arctan b/a . So I somewhat got stuck in a loop.

Q2: We can't use the auxiliary equation because a is also complex. So it doesn't work. I tried expanding both z and a only to get stuck eventually in an overly complicated equation. Any suggestions on just how to start it?

r.a.c. said:
I actually have two questions, both are in the complex plane.

Homework Statement

Q1: Express 962 as a sum of two squares (Hint: 962 = (13)(74)
Q2. Given z,a belong to C (complex). Find a such that the roots of the equation $$z^2 + az + 1 = 0$$ have equal absolute values (or modulus).

Homework Equations

Well this is about the complex plane so for the first one I think most probably its Euler's Identity. As for the second it could be anything.

The Attempt at a Solution

Q1: It is a sum of squares so the best I can think of is that 962 is a modulus of a vector i.e. z belongs to C, |z| = 962. Because z = a + ib so $$962^2 = a^2 + b^2$$. Then I try writing z in polar so we have $$z = |z|e^i^x$$ , where x is the argument of z. But x = arctan b/a . So I somewhat got stuck in a loop.
The problem asks you to find a and b such that $a^2+ b^2= 962$, NOT $962^2$! Although it doesn't have anything to do with this hint, one way to do it is by "brute strength": Is 962- 1 a square? 962- 4? 962- 9?...

Q2: We can't use the auxiliary equation because a is also complex.
I have no idea what you mean by this. What "auxiliary equation? And the real numbers are a subset of the complex numbers- saying that "a is a complex number" doesn't mean it can't be real. There is an obvious real number solution for a.

So it doesn't work. I tried expanding both z and a only to get stuck eventually in an overly complicated equation. Any suggestions on just how to start it?

The auxiliary equation is $$x = \frac{-b \frac{+}{-} \sqrt{4ac - b^2}}{2a}$$ where $$ax^2+bx+c=0$$. That's what I know it as anyways. Anyways, even the teacher said you can't do it that way. But if you can find a way, then I'd be more than happy.

You're right, I'm not saying a can't be real, but we also cannot make the assumption that it is purely real or imaginary. Why is there an obvious real number solution?

As for the sum of squares: I tried by brute strength, which did work, but its time consuming and has nothing to do with the hint or what we're studying. It turned out to be $$962 = 11^2+29^2$$.

Last edited:

1. What is the formula for finding the sum of two squares?

The formula for finding the sum of two squares is (a + b)^2 = a^2 + 2ab + b^2.

2. How do complex numbers factor into the sum of two squares?

Complex numbers can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit. When finding the sum of two squares involving complex numbers, we can use the formula (a + bi)^2 = a^2 - b^2 + 2abi.

3. Can the sum of two squares ever be a complex number?

No, the sum of two squares will always result in a real number. This is because the square of any real or complex number will always be a real number.

4. How do we solve for the sum of two squares when given a complex quadratic equation?

To solve for the sum of two squares in a complex quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. The values of a, b, and c can be found by rearranging the equation into standard form: ax^2 + bx + c = 0.

5. Are there any real-life applications of the sum of two squares and complex quadratic equations?

Yes, the sum of two squares and complex quadratic equations are used in various fields such as physics, engineering, and economics. They can be used to model and solve real-world problems involving complex systems and relationships.

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