The Equation:
$$ z^{4} -2 z^{3} + 12z^{2} -14z + 35 = 0 $$
has a root with a real part 1, solve the equation.
When it says a real part of 1, does this mean that we could use (z-1) and use polynomial division to extract the other rots? Hence:
$$ (z^{4} -2 z^{3} + 12z^{2} -14z + 35 ) / (z-1) $$
But i don't get the right answer, any other approaches I should try?
On
Since all coefficients are real, and $P(1)\neq0$, this means that we have two complex conjugate roots of the form $1\pm ai$, whose resultant quadratic is $z^2-sz+p$, where s and p are the sum and the product of the two roots. Can you take it from here ? ;-)
On
This is not so different from Lucian's approach, and may (or may not) feel computationally more straightforward.
This equation has real coefficients so that roots come in complex conjugate pairs. We are give that there are roots $z=1\pm bi$. Now let $y=z-1$ so that $z=y+1$ whence $$(y+1)^4-2(y+1)^3+12(y+1)^2-14(y+1)+35=y^4+2y^3+12y^2+8y+32=0$$ has the purely imaginary roots $y=\pm bi$
Putting this in and equating real and imaginary parts gives $$b^4-12b^2+32=0$$ and $$-2b^3+8b=0$$
The first equation factorises as $(b^2-4)(b^2-8)=0$ and the second as $-2b(b^2-4)=0$ so to be consistent we need $b^2=4$, $y=\pm 2i$ and $z=1\pm 2i$.
This gives the factorisation $$(z^2-2z+5)(z^2+7)=0$$ for the original equation, which enables the remaining roots to be found.
If there's a root of that equation of the form $1+ib$ then:
$$(1+ib)^4 - 2(1+ib)^3 -12(1+ib)^2-14(1+ib) + 35 = 0\Longrightarrow \\(1 + 4ib -6b^2 -4ib^3 + b^4) - 2(1 + 3ib - 3b^2 -ib^3) + 12(1 + 2ib - b^2) - 14(1+ib) + 35 =0 \Longrightarrow b^4-12b^2+32 + i(-2b^3+8b) = 0$$
Therefore:
$$\begin{cases}b^4 - 12b^2+32 = 0\\-2b^3+8b = 0 \end{cases}$$
From the second equation we can see that $b\in \lbrace -2,0,2\rbrace$.
But $b=0$ doesn't satisfy the first equation. The other two values do.
Therefore $(1-2i)$ and $(1+2i)$ are roots of the equation.