Problem source

Let  $x+{\rm i} y$ be a root of the quadratic equation $z^2 + pz +1=0$, where $p$ is a real number. Show that $x^2-y^2 +px+1=0$ and $(2x+p)y=0$.
Show further that either $p=-2x$ or $p=-(x^2+1)/x$ with $x\ne0$.
Hence show that the set of  points in the Argand diagram that can (as $p$ varies) represent roots of the quadratic equation consists of the real axis with one point missing and a circle.
This set of points is called the \textit{root locus} of the 
quadratic equation.
Obtain and sketch in the Argand diagram the root locus of the equation
\[ pz^2 +z+1=0\, \]
and the root locus of the equation
\[ pz^2 + p^2z +2=0\,.\]

Solution source

\begin{align*}
&& 0 &= z^2 + pz + 1\\
&&&= (x+iy)^2 + (x+iy)p + 1 \\
&&& = (x^2-y^2+px+1) + (2xy+py)i \\
\Rightarrow && 0 &= x^2 - y^2 + px + 1 \\
&& 0 &= (2x+p)y \\
\Rightarrow && p &= -2x \\
\text{ or } && y &= 0 \\
\Rightarrow && p &= -(x^2+1)/x
\end{align*}

Therefore as $p$ varies with either have $y = 0$ and $x$ taking any real value except $0$ ie the real axis minus the origin.

Or $p = -2x$ and $-y^2-x^2+1 = 0 \Rightarrow x^2 + y^2 = 1$ which is a circle.

Suppose $pz^2 + z + 1 = 0$

\begin{align*}
&& 0 &= pz^2 + z +1\\
&&&= p(x+iy)^2 + (x+iy) + 1\\
&&&= (px^2-py^2+x+1) + (2xyp + y) i \\
\Rightarrow && 0 &= (2xp+1)y \\
\Rightarrow && y & = 0, p = \frac{-(x+1)}{x^2}, x \neq 0 \\
\text{ or } && p &= -\frac{1}{2x}\\
\Rightarrow && 0 &= -\frac{1}{2}x + \frac{y^2}{2x} + x + 1 \\
&&&= \frac{y^2 - x^2 +2x^2 + 2x}{2x} \\
&&&= \frac{(x+1)^2+y^2-1}{2x}
\end{align*}

So we either have the real axis (except $0$) or a circle radius $1$ centre $(-1, 0)$ (excluding $x = 0$).




\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/(1+(#1)*(#1))};
    \def\xl{-2.5}; 
    \def\xu{2.5};
    \def\yl{-2.5}; \def\yu{2.5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid (\xu,\yu);
    

    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$\textrm{Re}(z)$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$\textrm{Im}(z)$};
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        \draw[curveA] (\xl, 0) -- (\xu, 0);
        \draw[curveA] (-1, 0) circle (1);
        \filldraw[white] (0,0) circle (1.5pt);
        \draw[curveA] (0,0) circle (1.5pt);
    \end{scope}
        % Set up axes


    \end{tikzpicture}
\end{center}

Suppose $pz^2 + p^2 z + 2 = 0$ then

\begin{align*}
&& 0 &= p(x+iy)^2 + p^2(x+iy) + 2 \\
&&&= (p(x^2-y^2) + p^2x + 2) + (2xyp + p^2y)i \\
\Rightarrow && 0 &= py(2x+p) \\
\Rightarrow && y &= 0, \Delta = x^4-8x \\
\Rightarrow && x &\in (-\infty, 0) \cup [2, \infty) \\
\text{ or } && p &= -2x \\
&& 0 &= (-2x)(x^2-y^2) + 4x^3+2 \\
&&&= 2x^3+2xy^2+2 \\
\Rightarrow && 0 &= x^3+xy^2+1
\end{align*}



\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/(1+(#1)*(#1))};
    \def\xl{-4}; 
    \def\xu{4};
    \def\yl{-4}; \def\yu{4};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid (\xu,\yu);
    

    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$\textrm{Re}(z)$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$\textrm{Im}(z)$};
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        \draw[curveA] (\xl, 0) -- (0, 0);
        \draw[curveA] (2, 0) -- (\xu, 0);
        % \draw[curveA] (-1, 0) circle (1);
        \filldraw[white] (0,0) circle (1.5pt);
        \draw[curveA] (0,0) circle (1.5pt);
        \filldraw[curveA] (2,0) circle (1.5pt) node[below, black] {$2$};
        \draw[curveA] (2,0) circle (1.5pt);

        \draw[curveA, domain=-1:-0.01, samples=150] plot({\x}, {sqrt((-1-\x^3)/\x)});
        \draw[curveA, domain=-1:-0.01, samples=150] plot({\x}, {-sqrt((-1-\x^3)/\x)});
    \end{scope}
        \node[below] at (-1, 0) {$-1$};
        % Set up axes


    \end{tikzpicture}
\end{center}