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For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with the matrix $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,$$ that squares to unity: $X^2=I$. Hence the spectrum of $A$ has $\pm$ symmetry: $$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ so if $\lambda$ is an eigenvalue then also $-\lambda$.

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity: $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,\;\;X^2=I.$$ Hence the spectrum of $A$ has $\pm$ symmetry: $$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ so if $\lambda$ is an eigenvalue then also $-\lambda$.

For real $a$, $b$, $c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with the matrix $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right)$$ that squares to unity. Hence the spectrum has $\pm$ symmetry.

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with the matrix $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,$$ that squares to unity: $X^2=I$. Hence the spectrum of $A$ has $\pm$ symmetry: $$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ so if $\lambda$ is an eigenvalue then also $-\lambda$.

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with the matri $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right)$$ that squares to unity. Hence the spectrum has $\pm$ symmetry.

For real $a$, $b$, $c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with the matrix $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right)$$ that squares to unity. Hence the spectrum has $\pm$ symmetry.