How can I resolve this equation: $z^4 + 2z^3 + 7z^2 − 18z + 26 = 0$, where there is a root that it's $1+ i$

Question

I know that if one root is $1+i$, other root is $1-i$.

But I don't know how I can find the last $2$ roots.

If you could explain in much detail why I always get lost in the little things, and sorry for my English.

Thanks for your help.

Answer 1

Since $\bigl(z-(1+i)\bigr)\bigl(z-(1-i)\bigr)=z^2-2z+2$ and since, by long division, you get that$$\frac{z^4+2z^3+7z^2-18z+26}{z^2-2z+2}=z^2+4z+13,$$the remaining roots are the roots of $z^2+4z+13$.

Answer 2

$(z-(1+i))(z-(1-i))=z^{2}-2z+2$. So factor the given polynomial in then form $(z^{2}-2z+2) (z^{2}+az+b)$. You can find $a$ and $b$ by comparing coefficients. Then solve the quadratic $z^{2}+az+b=0$. [ You should get $a=4$ and $b=13$].

Answer 3

Since the polynomial has real coefficients, If $1+i$ is a root, then its complex conjugate $1-i$ is also a root. In particular, then, the polynomial is divisible by $$(z-(1+i))(z-(1-i)) = z^2 - 2z + 2$$

Now either do the polynomial division, or equivalently solve $$z^4+2z^3+7z^2−18z+26=(z^2 -2z +2)(z^2 + ax + b)$$

By expanding the right side and comparing the coefficients of the two 4th-degree polynomials, you get $a= 4, b=13$.

The last step is to solve $$z^2 +4z +13 = 0$$

which has solutions $z=-2 \pm 3i$.

Answer 4

Since $1+i+1-i=2$ and $(1+i)(1-i)=2$, we got a factor $z^2-2z+2.$

Now, use $$z^4 + 2z^3 + 7z^2 − 18z + 26=$$ $$=z^4-2z^3+2z^2+4z^3-8z^2+8z+13z^2-26z+26=$$ $$=(z^2-2z+2)(z^2+4z+13)=((z-1)^2+1)((z+2)^2+9).$$ Can you end it now?

Answer 5

The best and the easiest way among all answers currently present.

The equation $z^4 + 2z^3 + 7z^2 − 18z + 26 = 0$

We know the roots, $1+i$ and $1-i$, we are sure that the product of the remaining $2$ roots is $13$. (using product of all roots =$\small{\text{constant}/\text{coef first term}}$ for even powers).

Also, the sum of all roots by Vietta's theorem is $-2$. So, the remaining two roots should sum up to $-4$ , $\implies Re(z_{remaining})=-2$ (since they exist in conjugate pairs) and $|z|=\sqrt{13}$.

Hence we get the roots as $-2+3i$ and $-2-3i$.