# Thread: Complex Number roots Easy question!

1. ## Complex Number roots Easy question!

Hi I have this question that is very easy I am sure but I am having a hard time trying to figure out what the last part is asking! anyways here is the question:

Find the four roots of $\displaystyle z^4 + 4 = 0$ and use them to factor $\displaystyle z^4 + 4$ into the product of two quadratic factors with real coefficients.

Ok I was able to find the 4 roots with ease they are $\displaystyle (1+i), (-1+i), (1-i), (-1-i)$

Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well

My GUESS was $\displaystyle (z^2-2z+2)(z^2+2z+2)$ but I didnt use the roots at all to come to that answer I used some formula I found on web.

So please someone out there make me less dumb.

2. $\displaystyle z^4 + 4 = 0$

$\displaystyle z^4 = -4$

$\displaystyle z^4 = 4\left(\cos{\pi} + i\sin{\pi}\right)$

$\displaystyle z = 4^{\frac{1}{4}}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$

$\displaystyle z = \sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$

Now it's important to note that all the roots are going to be evenly spaced around a circle, so they will all have a magnitude of $\displaystyle \sqrt{2}$ and differ by angles of $\displaystyle \frac{2\pi}{4} = \frac{\pi}{2}$.

Find the other roots.

3. did you read my entire post? I'm having trouble with the 2nd part of question. that has to do with product of two quadratic factors with real coefficients.

4. Originally Posted by fizzle45
Hi I have this question that is very easy I am sure but I am having a hard time trying to figure out what the last part is asking! anyways here is the question:

Find the four roots of $\displaystyle z^4 + 4 = 0$ and use them to factor $\displaystyle z^4 + 4$ into the product of two quadratic factors with real coefficients.

Ok I was able to find the 4 roots with ease they are $\displaystyle (1+i), (-1+i), (1-i), (-1-i)$

Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well

My GUESS was $\displaystyle (z^2-2z+2)(z^2+2z+2)$ but I didnt use the roots at all to come to that answer I used some formula I found on web.

So please someone out there make me less dumb.
Note that $\displaystyle (z-(1+i))(z-(1-i)) = z^2-2z+2$

5. Originally Posted by fizzle45
Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well
If $\displaystyle (x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0$, the coefficients of $\displaystyle x$ in $\displaystyle f(x)$ are are $\displaystyle a_{0}, a_{1}, a_{2}, ..., a_{n}$, and could be real or complex numbers. For example, if we have $\displaystyle p(x) = 3x^2+5x+7$ and $\displaystyle q(x) = 5ix^4+ix$, then all the coefficients of $\displaystyle x$ in $\displaystyle p(x)$ are real (i.e., $\displaystyle 3, 5, 7$ are real) but the coefficients of $\displaystyle q(x)$ are not (since $\displaystyle 5i$ and $\displaystyle i$ are complex). When they ask you to provide two quadratic factors with real coefficients, they want two quadratic functions whose coefficients are not complex numbers. If you have numbers $\displaystyle n_{1},...., n_{k}$, then the minimal polynomial for which all these numbers satisfy is $\displaystyle (x-n_{1})\cdots(x-n_{k})$. Take $\displaystyle 1, 2, 3$. To find the polynomial whose roots are these, we find the product $\displaystyle (x-1)(x-2)(x-3) = (x^2-3x+2)(x-3) = x^3-6 x^2+11 x-6$. Note that if we were asked to write it as a product of a linear and a quadratic, we would have just stopped at the second step. What you were asked is bit reversed but essentially the same: given the roots of $\displaystyle f(x) = x^4+4 = 0$ are $\displaystyle a, b, c,$ and $\displaystyle d$ (all of which you have found), write $\displaystyle f(x)$ as a product of two quadratics with real coefficients.

6. The point of this exercise is to show you about conjugate pairs as roots of polynomials with real coefficients. If $\displaystyle w$ is a complex number then $\displaystyle \left( {z - w} \right)\left( {z - \overline w } \right) = z^2 - \left( {w + \overline w } \right)z + w\overline w = z^2 - 2\text{Re} (w)z + \left| w \right|^2$.
That is a polynomial with real coefficients.

7. Hello, fizzle45!

Find the four roots of $\displaystyle z^4 + 4 = 0$

and use them to factor $\displaystyle z^4 + 4$ into the product of two quadratic factors
with real coefficients.

Ok, I was able to find the 4 roots with ease.
They are: $\displaystyle (1+i), (1-i), (\text{-}1+i), (\text{-}1-i)$ .Good!

Now for the part that asks product of two quadratic factors with real coefficients,
I have no idea what they are even asking

My GUESS was: $\displaystyle (z^2-2z+2)(z^2+2z+2)$ . Right!
but I didnt use the roots at all to come to that answer.
I used some formula I found on web.

If a polynomial equation, $\displaystyle f(z) = 0$,has four roots: .$\displaystyle a,b,c,d$
. . then the equation factors into: .$\displaystyle (z-a)(z-b)(z-c)(z-d) \:=\:0$

We have the equaton: .$\displaystyle z^4 + 4 \:=\:0$
. . with four roots: .$\displaystyle (1+i),\:(1-i),\;(\text{-}1+i),\;(\text{-}1-i)$

So the equation factors:

. . $\displaystyle \bigg[z - (1+i)\bigg]\,\bigg[z - (1-i)\bigg]\,\bigg[z-(-1+i)\bigg]\bigg[z-(-1-i)\bigg] \;=\;0$

. . . . .$\displaystyle \underbrace{\bigg[(z-1) - i\bigg]\,\bigg[(z-1) +i\bigg]}_{\text{multiply}}\,\underbrace{\bigg[(z+1) - i\bigg]\,\bigg[(z+1) + i\bigg]}_{\text{multiply}} \;=\;0$

The product of the first two factors is:
. . $\displaystyle (z-1)^2 - i^2 \:=\:z^2 - 2z + 1 + 1 \:=\:z^2-2z + 2$

The product of the last two factors is:
. . $\displaystyle (z+1)^2 - i^2 \:=\:z^2 + 2z + 1 + 1 \:=\:z^2 + 2z + 2$

And that is why: .$\displaystyle z^4 + 4 \;=\;(z^2 - 2z + 2)(z^2 + 2z + 2)$