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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.

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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 $b$.

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Thank you so much! Jeez, this homework is really really hard, don't suppose you'd fancy helping on some of the other questions? – user2930356 Jan 23 '14 at 22:19
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. – Old John Jan 23 '14 at 22:21
Nice John and thanks for your comment at FB. Babak +1 – Babak S. 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 $$(x-z)(x-\overline{z})=(x^2-2\operatorname{Re}(z)+|z|^2)$$

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Hope you Angel gets better. I am thinking of you. :+) – Babak S. 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.

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