In search for the purely imaginary roots of the polynomial $p(z) = z^4 + z^3 + 4z^2 + 3z + 7$, I first replaced $z$ by $iy$ for some $y \in \mathbb{R}$ which yields
$$p(iy) = y^4 – i y^3 – 4 y^2 + 3 iy + 7$$
which by separating into Real and Imaginary parts yields
$$p(iy) =( y^4 – 4 y^2 + 7 ) + iy (3-y^2) .$$
To find the common zeros let $\text{Re} f(z)=0$ and $\text{Im} f(z) =0$ which yields respectively,
$$y=2 \pm i\sqrt{3} ~~ \text{and} ~~ y =0, \pm \sqrt{3} .$$
The first pair of values turned out to be themselves complex numbers. So I guess there are no purely imaginary roots as $y$ supposed to be a real number.
Next, I want to find the change of the argument of $\text{arg} (f(z))$ when $z$ traverses on the imaginary axis from $z = R$ (sufficiently large) to $z=0$. How can I go about and solving this problem ? Any help is much appreciated.
Best Answer
$y$ must be a real number.
Since there is no real number satisfying $y^4-4y^2+7=0$ and $y(3-y^2)=0$ at the same time, there is no purely imaginary root of $z^4 + z^3 + 4z^2+3z+7 = 0$.