# How do we solve a polynomial where all roots are imaginary(complex)? [closed]

I had a problem with a specific question from my textbook. This is the question: Prove that $1+i$ is a root of the equation $z^4 + 3z^2 - 6z + 10 = 0$. Find all other roots.

Answers: $1-i$, $-1+2i$, $-1-2i$

## closed as off-topic by Namaste, Vidyanshu Mishra, Aweygan, John B, kingW3Oct 11 '17 at 17:04

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• Did you try to replace $z=1+i$ into the polynomial? – IEDC PHY Oct 11 '17 at 16:15
• if $1+i$ is a root, than $1-i$ is also a root – Dr. Sonnhard Graubner Oct 11 '17 at 16:16
• Well, the coefficients of the polynomial are real, so the roots come in complex conjugate pairs. Each root and its conjugate are together the roots of a quadratic expression which is a factor of the original quartic. You have one root, find the relevant quadratic, divide though to find the other quadratic factor and find the roots of that quadratic in the usual way. – Mark Bennet Oct 11 '17 at 16:17
• Well they didn't teach that in school. Thank you so much. – Master Boggins Oct 11 '17 at 16:21

Guide:

Notice that all the coefficients are real. Hence if $1+i$ is a root, its conjugate is a root of the polynomial as well. From there we can form a quadratic equation $p_1$ with these roots as the root.

We can denote then divide $z^4+3z^2-6z+10$ by $p_1$ to obtain another quadratic equation and solve it.

• Thank you so much, that was very helpful. – Master Boggins Oct 11 '17 at 16:24

If $1+i$ is a root then since our polynomial has real coefficients, we see that also $1-i$ should be a root of the equation.

Thus, our polynomial should be divisible by $z^2-2z+2$, which gives the following trying: $$z^4+3z^2-6z+10=$$ $$=z^4-2z^3+2z^2+2z^3-4z^2+4z+5z^2-10z+10=$$ $$=(z^2-2z+2)(z^2+2z+5)=((z-1)^2+1)(z+1)^2+4)=$$ $$(z-(1+i))(z-(1-i))(z-(-1+2i))(z-(-1-2i)).$$

Now, we see that $1+i$ is indeed the root and we got the answer: $$\{1+i,1-i,-1+2i,-1-2i\}$$

• Sorry, that just isn't helpful at all. – Thomas Andrews Oct 11 '17 at 16:21
• Thanks for the detailed steps, sir. The fact that fators are conjugate pairs if the coefficients are real is all I needed to know. – Master Boggins Oct 11 '17 at 16:29
• @Master Boggins You are welcome! – Michael Rozenberg Oct 11 '17 at 16:31
• @Thomas Andrews I ended to write my post. See now. – Michael Rozenberg Oct 11 '17 at 16:32