\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(x + iy\) is a root of this quadratic equation, putting it back into the original equation, we have \[ (x + iy)^2 + p (x + iy) + 1 = (x^2 - y^2 + px + 1) + (2x + p)yi = 0, \] and so it must have both real parts and complex parts \(0\), and hence \(x^2 - y^2 + px + 1 = 0\), and \((2x + p)y = 0\).
Since \((2x + p)y = 0\), we must have either \(2x + p = 0\) (which gives \(p = -2x\)), or \(y = 0\). In the latter case, we put this back into the first equation, and we have \[ x^2 + px + 1 = 0. \]
If \(x = 0\), then we must have \(0 + 0 + 1 = 1 = 0\) which is impossible. Hence, \(x \neq 0\), and by rearranging, we have \[ p = - \frac {x^2 + 1}{x}. \]
In the case where \(p = -2x\), we must have \[ x^2 - y^2 + (-2x) \cdot x + 1 = 0 \iff x^2 + y^2 = 1, \] and this represents a circle centred at the origin with radius \(1\).
In the case where \(p = - \frac {x^2 + 1}{x}\), we must have \(y = 0\), and \(x \neq 0\). This represents the real axis without the origin.
This is the root locus of this equation.
For the second equation, let \(z = x + iy\) be a solution. We have \[ p (x + iy)^2 + (x + iy) + 1 = (px^2 - py^2 + x + 1) + (2px + 1)yi = 0, \] and so \(px^2 - py^2 + x + 1 = 0\) and \((2px + 1)y = 0\).
Since \((2px + 1)y = 0\), we must have either \(2px + 1 = 0\) (which gives \(p = - \frac {1}{2x}\) since \(x \neq 0\), or otherwise \(0 + 1 = 1 = 0\)), or \(y = 0\). In the latter case, we put this back to the first equation, and we have \[ px^2 + x + 1 = 0. \]
If \(x = 0\) then we must have \(0 + 0 + 1 = 1 = 0\) which is impossible. Hence, \(x \neq 0\), and by rearranging, we have \[ p = - \frac {x + 1}{x^2}. \]
In the case where \(p = - \frac {1}{2x}\), given \(x \neq 0\), \[ - \frac {1}{2x} (x^2 - y^2) + x + 1 = 0 \iff \frac {x}{2} + \frac {y^2}{2x} + 1 = 0 \iff (x + 1)^2 + y^2 = 1. \]
This represents a circle centred at \((-1, 0)\) with radius \(1\), and since \(x \neq 0\), we have to remove the point \((0, 0)\).
In the case where \(p = -\frac {x + 1}{x^2}\), \(y = 0\) and this represents the real axis without the origin.
This is the root locus of this equation.
For the final equation, let \(z = x + iy\) be a solution. We have \[ p(x + iy)^2 + p^2 (x + iy) + 2 = (px^2 - py^2 + p^2x + 2) + yp(2x + p)i = 0, \] and so \(px^2 - py^2 + p^2x + 2 = 0\) and \(yp(2x + p)\) = 0.
Notice that here, \(p \neq 0\), since if \(p = 0\) then \(2 = 0\) and there is no solution. So since \(yp (2x + p) = 0\), we have \(2x + p = 0\) which gives \(p = -2x\), or \(y = 0\). In the latter case, we put this back to the first equation, and we have \[ p x^2 + p^2 x + 2 = 0. \]
If \(x = 0\) then we must have \(0 + 0 + 2 = 2 = 0\) which is impossible. Hence, \(x \neq 0\). For this to have a real solution for \(p\), we must have \(x \neq 0\) and \[ (x^2)^2 - 4 \cdot x \cdot 2 \geq 0, \] which means \[ x (x - 2) (x^2 + 2x + 2) \geq 0. \]
Since \(x^2 + 2x + 2 = (x + 1)^2 + 1 \geq 1 \geq 0\), we must have \(x (x - 2) \geq 0\), and \(x \leq 0\) or \(x \geq 2\). This represents the real line with the interval \([0, 2)\) removed.
In the case where \(p = -2x\), putting this back to the first equation, we have \[ (-2x) x^2 - (-2x)y^2 + (-2x)^2 x + 2 = 0 \iff x^3 + xy^2 + 1 = 0 \iff y^2 = -\frac {1 + x^3}{x}. \]
This is the root locus of this equation.
