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I have an equation that I don't know how to solve.

The homework assignment is as follows:

$z^4-4z^3+ \frac{5}{4}z^2-z+ \frac{1}{4} = 0$ that has a solution $z=\frac{1}{2}i$

Solve the equation completely.

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closed as off-topic by Care Bear, Mark Fantini, Bryan, RecklessReckoner, Yiorgos S. Smyrlis Sep 1 at 18:06

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5  
You're given a non-null solution. Since it's a real polynomial, this immediately gives you a different solution. You can then factor the given polynomial as the product of two real polynomials. –  Git Gud Sep 1 at 14:45
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The coefficients are real therefore $z=-\frac{i}{2}$ is a root. Therefore you can divide that polynomial by $z^2+\frac{1}{4}$ –  rlartiga Sep 1 at 14:46

5 Answers 5

HINT : Using the complex conjugate root theorem, we know that $z=-i/2$ is also a solution. Hence, we can divide LHS by $$\left(z+\frac i2\right)\left(z-\frac i2\right)=z^2+\frac 14.$$

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Hint:you basically have

$z^4-4z^3+z^2+(1/4)z^2-z+(1/4)$ $=z^2(z^2-4z+1)+(1/4)(z^2-4z+1)$

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1  
It is not completely obious to me how you arrive at your equation. Is it an (educated) guess. Did you find it by chance? Which steps did you actually take? –  Bernhard Sep 1 at 18:01
    
It is like you said a mere guess, but It is usually the case in such questions where the equation has nice solutions, that it expected that you will get such results, so you know what you should look for. –  Avrham Aton Sep 1 at 19:15

Here is a workmanlike way of proceeding:

Multiply by $4$ to obtain $4z^4-16z^3+5z^2-4z+1=0$ (not a necessary step, but I like clearing fractions)

Note the quadratic factor $(4z^2+1)$ implied by the root you are given

Note the leading and constant terms to obtain the partial factorisation $$(4z^2+1)(z^2+az+1)$$where $a$ is unknown.

Determine $a$ by comparing coefficients of $z$ to obtain $-4=a$

Find the roots of the remaining factor: $z^2-4z+1=0$

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1  
"Determine $a$ by comparing coefficients of $z$" -- this is what happens when schools don't teach long division ;-p –  Steve Jessop Sep 1 at 16:36
    
@SteveJessop: And this is what happens when schools don't teach manual extraction of square roots. Humanity is doomed. Doomed, I tell you. –  TonyK Sep 1 at 19:27
    
Anything to avoid long division, especially when you know the remainder is zero, and have only one piece of information to find. –  Mark Bennet Sep 1 at 19:29

I have a slightly different approach you may like to consider:

We know that $z^4-4z^3+\frac{5}{4}z^2 - z + \frac{1}{4}=0$ can be written in the form $(z-a)(z-b)(z-c)(z-d)=0$ where $a,b,c,d$ are the roots. We already know one, so we have $a= \frac{1}{2}i$, and as pointed out by others, we know that the conjugate of $a$ is also a root because the polynomial has real coefficients. Thus, we can say $b=-\frac{1}{2}i$.

So far we have $(z-\frac{1}{2}i)(z+\frac{1}{2}i)(z-c)(z-d)=0 \Rightarrow (z^2+\frac{1}{4})(z-c)(z-d)=0$. Now since you know that this factored form of the polynomial must be equivalent to the original one we started with, we expand it and compare the coefficients with the original equation. So, $$(z^2+\frac{1}{4})(z-c)(z-d)=z^4+(-c-d)z^3+(cd+\frac{1}{4})z^2+(\frac{-c-d}{4})z + \frac{cd}{4}$$ By comparing the coefficients we get a system of equations $$\frac{-c-d}{4}=-1 \\ \frac{cd}{4}=\frac{1}{4}$$ The solution of the system is $c=2-\sqrt{3}$ and $d=2+\sqrt{3}$

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$z^4-4z^3+\frac{5}{4}z^2-z+\frac{1}{4}=0 \Leftrightarrow z^4-4z^3+z^2+\frac{1}{4}z^2-z+\frac{1}{4}=0$

Now use factoring by grouping:

$z^2(z^2-4z+1)+\frac{1}{4}(z^2-4+1)=0$

$(z^2-\frac{1}{4})(z^2-4z+1)=0$

I am sure you can take it from here.

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