Write in the form $f(z) = 0$, where $f(z)$ is a polynomial of degree $4$ with real coefficients, the equation having $(3 + i)$ and $(1 + 3i)$ as two of its roots.

Can anyone help me? I'm guessing the two other roots are $(3-i)$ and $(1-3i)$ as they are the complex conjugates of the original roots.

OKAY Thank you for your response, I understand the question and the answer now and I will use that conjugate theorem lots from now on.


Find the real root of the equation $z^3 + z + 10 = 0$ given that one complex root is $1 – 2i$.

I've realised that the roots are $(1-2i), (1+2i)$, and a real number we'll call a

So using the theorem got me $(z-1-2i)(z-1+2i)(z-x)$

No idea on where to go next.


Your reasoning is fine, and you can get a polynomial with these roots by taking the product of factors $z-\alpha$ where $\alpha$ is the root(s):

$$(z-3-i)(z-3+i)(z-1-3i)(z-1+3i) = 0$$

You can then simplify, either by multiplying out the brackets, in pairs, as written above, to keep the algebra straightforward:

$$(z-3-i)(z-3+i)(z-1-3i)(z-1+3i) = [(z-3)^2+1][(z-1)^2+9]\\= (z^2-6z+10)(z^2-2z+10), \text{etc.}$$

A really useful identity to remember in these situations is:

$$(a+ib)(a-ib) = a^2 + b^2$$

with suitable choices for $a$ and $b$.

  • $\begingroup$ Thank you so much! Jeez, this homework is really really hard, don't suppose you'd fancy helping on some of the other questions? $\endgroup$ Jan 23 '14 at 22:19
  • 2
    $\begingroup$ You can try posting some others, but remember always to explain what you have tried, or how far you have got, and exactly where you are stuck. I may not be online, but others will help. $\endgroup$
    – Old John
    Jan 23 '14 at 22:21
  • $\begingroup$ Nice John and thanks for your comment at FB. Babak +1 $\endgroup$
    – Mikasa
    Jan 25 '14 at 18:03

Let denote $$P(x)=ax^4+bx^3+cx^2+dx+e$$ the desired polynomial so if $z$ is a complex root of $P$ then since the coefficients are real $$\overline{P(z)}=a\overline{z}^4+b\overline{z}^3+c\overline{z}^2+d\overline{z}+e=0$$ so $\overline{z}$ is also a root of $P$.

Finaly to figure out the polynomial you need this equality $$(x-z)(x-\overline{z})=(x^2-2\operatorname{Re}(z)+|z|^2)$$

  • $\begingroup$ Hope you Angel gets better. I am thinking of you. :+) $\endgroup$
    – Mikasa
    Jan 26 '14 at 6:58

If you say that a particular number is a root, that means $x$ minus that number must be a factor. So $(x-(3+i))$ and $(x-(1+3i))$ must be factors.

You say it has real coefficients. That implies if $(x-(3+i))$ is a factor, then $(x-(3-i))$ must also be a factor, and if $(x-(i+3))$ is a factor, then $(x-(i-3))$ must be a factor. The two numbers $3\pm i$ are each other's complex conjugates and the two numbers $1\pm 3i$ are each others complex conjugates. Coming in complex-conjugate pairs is what can make the imaginary parts cancel when you muliply out the polynomial.

You now have four first-degree factors, so when you multiply them, you get a fourth-degree polynomial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.