Finding a Quadratic Factor of z⁴+16

[XtremePhysX]

Homework Statement

Find a real quadratic factor of the polynomial:

Homework Equations

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

The Attempt at a Solution

I don't know :$

 

[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.

 

[oli4]

Hi XtremePhysX
maybe this link will be helpful:
Roots of unity

 

[XtremePhysX]

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

 

[Mentallic]

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

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

What does z equal to then?

 

[oli4]

Hmm sorry, My previous links was not that helpful.
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...

 

[uart]

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

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.

 

[XtremePhysX]

thanks guys

 

[XtremePhysX]

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

can I split that into linear factors?

 

[ehild]

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

can I split that into linear factors?

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

ehild

 

[XtremePhysX]

Thanks a lot ehild.

 

[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 z_k=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

 

[XtremePhysX]

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 z_k=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

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.

 

[Mentallic]

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.

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.

 

[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.

 

[eumyang]

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

So this is the solution.

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).

 

[Mentallic]

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

So this is the solution.

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.

Start by answering these:

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

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

 

[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

 

[XtremePhysX]

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.

Start by answering these:

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

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

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

 

[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!

 

[XtremePhysX]

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

Isn't √i either:
$$e^{\frac{i\pi }{4}}$$
or $$cis(\frac{\pi }{4})$$

 

[ehild]

Isn't √i either:
$$e^{\frac{i\pi }{4}}$$
or $$cis(\frac{\pi }{4})$$

Yes, but what is it in numerical form, a+bi without cis and pi?
The linear factors of z⁴+16 can be written in the form (z-2(a+bi))=z-2a-2bi. Find all these factors.

ehild

 

[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.

Hi XtremePhysX. I posted step by step instructions of the easiest way to do this in my previous post. If you're unsure of any of the steps just ask.

You can get the 4 complex roots like this:

$$z^4 = 16 e^{i(\pi + 2k \pi)}$$

$$z = 2 e^{i(\frac{\pi}{4} + \frac{k \pi}{2})}$$

Substituting any 4 consecutive integers for k (eg k=-2,-1,0,1) gives,

$$z_1 = 2 e^{\frac{-3i\pi}{4}}$$

$$z_2 = 2 e^{\frac{-i\pi}{4}}$$

$$z_3 = 2 e^{\frac{i\pi}{4}}$$

$$z_4 = 2 e^{\frac{3i\pi}{4}}$$

Now just follow the steps given above.

 

[Mentallic]

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

Right, and I see that many others here have explained what the roots of z⁴+16=0 are, so we will move onto the next step.

Since $cis(\pi/4)$ and its conjugate pair $cis(-\pi/4)$ as well as $cis(3\pi/4)$ and its conjugate pair $cis(-3\pi/4)$ are all roots of the quartic, we can express it as follows:

$$z^4+16=(z-cis(\pi/4))(z-cis(-\pi/4))(z-cis(3\pi/4))(z-cis(-3\pi/4))$$

Now, expand $(z-cis(\pi/4))(z-cis(-\pi/4))$ and use the results you've posted to turn it into a real quadratic factor.
Then do the same for $(z-cis(3\pi/4))(z-cis(-3\pi/4))$ and you're done

 

[XtremePhysX]

$$z^{2}+2z\sqrt{2}+4$$

 

[ehild]

$$z^{2}+2z\sqrt{2}+4$$

Yes, and the other one?

ehild

 

[ehild]

$$z^4+16=(z-2cis(\pi/4))(z-2cis(-\pi/4))(z-2cis(3\pi/4))(z-2cis(-3\pi/4))$$

You forgot "2" in front of cis.

ehild

 

[Mentallic]

You forgot "2" in front of cis.

ehild

haha thanks for spotting that

 

[XtremePhysX]

$$z^{2}+2z\sqrt{2}+4$$

The question says a real quadratic factor so I though that implies there should be only 1 factor. Isn't that right?

 

[Mentallic]

The question says a real quadratic factor so I though that implies there should be only 1 factor. Isn't that right?

Yes you're right, but I don't understand why they'd make you stop there.

 

[XtremePhysX]

Yes you're right, but I don't understand why they'd make you stop there.

It's a 2 mark question in an MX2 paper, I think you know what MX2 is ;)

 

[ehild]

The second one is obvious: The product of the other two linear factors: z²-2√2 z+4. Check if multiplying these two factors you really get z⁴ +16.

ehild

 

[XtremePhysX]

The second one is obvious: The product of the other two linear factors: z²-2√2 z+4. Check if multiplying these two factors you really get z⁴ +16.

ehild

I completely understand it now, thanks to everyone, much appreciated.

 

[Mentallic]

It's a 2 mark question in an MX2 paper, I think you know what MX2 is ;)

Oh, well then a lot of time would've been wasted finding the other complex roots as well if only one quadratic factor is necessary. Look at how quick and easy the solution can be:

$$z^4+16=0$$
$$z^4=-16$$
$$z^4=16cis(\pi+2k\pi)$$
$$z=2cis(\frac{\pi(1+2k)}{4})$$ for $k=-2,-1,0,1$

Now let's just consider one of the complex conjugate pairs:

$$z_1=2cis(\pi/4)$$
$$z_2=2cis(-\pi/4)=\bar{z_1}$$

Therefore the quadratic factor is,

$$(z-z_1)(z-\bar{z_1})$$
$$=z^2-(z_1+\bar{z_1})z+z_1\bar{z_1}$$
Where $$z_1+\bar{z_1}=2Re(z_1)=4cos(\pi/4)=2\sqrt{2}$$
and $$z_1\bar{z_1}=4|z_1|^2=4$$

Therefore we have the quadratic factor
$$z^2-2\sqrt{2}z+4$$

 

[XtremePhysX]

Oh, well then a lot of time would've been wasted finding the other complex roots as well if only one quadratic factor is necessary. Look at how quick and easy the solution can be:

$$z^4+16=0$$
$$z^4=-16$$
$$z^4=16cis(\pi+2k\pi)$$
$$z=2cis(\frac{\pi(1+2k)}{4})$$ for $k=-2,-1,0,1$

Now let's just consider one of the complex conjugate pairs:

$$z_1=2cis(\pi/4)$$
$$z_2=2cis(-\pi/4)=\bar{z_1}$$

Therefore the quadratic factor is,

$$(z-z_1)(z-\bar{z_1})$$
$$=z^2-(z_1+\bar{z_1})z+z_1\bar{z_1}$$
Where $$z_1+\bar{z_1}=2Re(z_1)=4cos(\pi/4)=2\sqrt{2}$$
and $$z_1\bar{z_1}=4|z_1|^2=4$$

Therefore we have the quadratic factor
$$z^2-2\sqrt{2}z+4$$

Neat solution, thank you.