# Homework Help: Complex Numbers

1. Aug 12, 2012

### XtremePhysX

1. The problem statement, all variables and given/known data

Find a real quadratic factor of the polynomial:

2. Relevant equations

$$z^{4}+16=0$$

3. The attempt at a solution

I don't know :\$

2. Aug 12, 2012

### Mentallic

There are a couple of ways of solving this problem.

If you want to solve it using complex numbers, then you need to find all the complex roots of $z^4=-16$ by converting -16 into mod-arg form.

If you want to solve it without, you'll instead be solving a system of equations to find the value of some unknown variables.

3. Aug 12, 2012

### oli4

Hi XtremePhysX
Roots of unity

4. Aug 12, 2012

### XtremePhysX

Thank you guys.
-16 = 16cis(pi)
where do I go from there?

5. Aug 12, 2012

### Mentallic

So $z^4=16cis(\pi)$

What does z equal to then?

6. Aug 12, 2012

### oli4

I think you might just want to solve it 'as it comes' by factorizing.
z⁴+16=0 -> (z²)²-(4i)²=0
so (z²-4i)(z²+4i)=0 ... (do the same trick once more and you will get your 4 roots easily)
Cheers...

7. Aug 12, 2012

### uart

The best way to solve it with complex numbers is to find the 4 complex roots.

1. Note that the roots can be grouped as pairs of complex conjugates.

2. We know that both the sum and product of complex conjugates are real numbers.

3. Use the relationship between the sum and product of the roots and the coefficients of a quadratic to find the factor.

8. Aug 12, 2012

### XtremePhysX

thanks guys

9. Aug 13, 2012

### XtremePhysX

so I got (z^2+4i)(z^2-4i)

can I split that into linear factors?

10. Aug 13, 2012

### ehild

yes, (z2-4i)=(z-2√(i))((z+2√(i)), and (z^2+4i)=(z-2√(-i))((z+2√(-i))....What is √i?

ehild

11. Aug 13, 2012

### XtremePhysX

Thanks a lot ehild.

12. Aug 13, 2012

### ehild

Have you solved the problem? What is the solution?

The more elegant method would have been to use the complex roots of -16, which are zk=2 cis[pi/4+k(2pi/4)], (k=0,1,2,3) and write z4=(z-2cis(pi/4))((z-2cis(3pi/4))((z-2cis(5pi/4))((z-2cis(7pi/4)). The product of the first and last factors is real and so is the product of the second and third ones.

ehild

13. Aug 13, 2012

### XtremePhysX

The question says this: Find a real quadratic factor of the polynomial $$z^{4}+16=0$$
Does that mean the factors can't be imaginary, I'm a bit confused here.

14. Aug 13, 2012

### Mentallic

No, real quadratic factors are factors that are quadratic (of the form $ax^2+bx+c$) where a,b,c are all real (not complex) coefficients.
All real quadratics with complex roots (or complex linear factors) have their roots that come in complex conjugate pairs, so that means that if we have two complex conjugate linear factors, say, $(x-(a+ib))$ and $(x-(a-ib))$ then we can always multiply them together to get back to a real quadratic.

15. Aug 13, 2012

### XtremePhysX

$$\left [ z+(2 (-1)^{1/4}) \right ] \left [ z-(2 (-1)^{1/4}) \right ] \left [ z+(2 (-1)^{3/4}) \right ] \left [ z-(2 (-1)^{3/4}) \right ]$$

So this is the solution.

16. Aug 13, 2012

### eumyang

Can you instead write the roots in standard form (a + bi)? ehild almost gave them to you when he said that the roots were cis(π/4), cis(3π/4), cis(5π/4), and cis(7π/4).

17. Aug 13, 2012

### Mentallic

Nowhere near. Those are linear factors, and they're not real factors either.
The question asks you to convert these complex linear factors into real quadratic factors.

$$cis(\pi/4)+cis(-\pi/4)=?$$

$$cis(\pi/4)cis(-\pi/4)=?$$

18. Aug 13, 2012

### ehild

Not yet, you can not use complex numbers. So substitute the roots of -1 in the form a+bi. Then multiply two factors which are complex conjugate to each other to eliminate the imaginary parts.
You know that √(-1)=±i. What is √i? What are the complex numbers which squares are i?

(hint: try cos(pi/4)+isin(pi/4))

ehild

19. Aug 13, 2012

### XtremePhysX

$$cis(\pi/4)+cis(-\pi/4)=\sqrt{2}$$
$$cis(\pi/4)cis(-\pi/4)=1$$

20. Aug 13, 2012

### Millennial

Protip: The square root of i can be derived from Euler's formula as follows: $e^{i\pi}=-1$, so it follows that $e^{i\pi/4}=\sqrt{i}$. However, we can convert this to the standard form using Euler's formula once again: $e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$
Using the same logic but instead using $e^{-i\pi}=-1$, one obtains the other square root of i.
Now plug these in and simplify!