Sum of two square and complex quadratic

[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?

 

[HallsofIvy]

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?

 

[r.a.c.]

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 $$962 = 11^2+29^2$$.