If $\frac{p}{p+q}$ is a negative real number, what can I deduce about complex $p$ and $q$? [closed]

Question

Let $p, q$ be complex numbers with non-negative real parts and arbitrary imaginary parts. If $\frac{p}{p+q}$ is a negative real number, what can I deduce about $p$ and $q$?

Motivation: This question comes up in analysis of AC circuits, where resistors have positive real impedance and other components exist which have arbitrary negative impedance. I'm looking to understand under what conditions the net impedance of a voltage divider can be negative, which should cause an inversion (180 degree phase shift)

Answer 1

Say $p/(p+q) = -a$ for some real $a>0$. Then $(1+a)p = -aq$. The left-hand side has non-negative real part, while the right-hand side has non-positive real part. This forces the real part of $p$ and $q$ to be both $0$, so $p = xi$ and $q = yi$, where $x, y \in \mathbb{R}$. Now, $p/(p+q) = x/(x+y) < 0 \implies x(x+y) < 0 \implies xy < -x^2$. If $x > 0$, this means $y < -x$, and if $x<0$ then $y>-x$.

Answer 2

Let $p = a + b i$ and $q = c + d i$ for some $a, b, c, d \in \mathbb {R}$. Then

$$\begin{align} \frac {p}{p + q} & = \frac {a + b i}{\left( a + c \right) + \left( b + d \right) i} \\ & = \frac {a + b i}{\left( a + c \right) + \left( b + d \right) i} \cdot \frac {\left( a + c \right) - \left( b + d \right) i}{\left( a + c \right) - \left( b + d \right) i} \\ & = \frac {\left[ a \left( a + c \right) + b \left( b + d \right) \right] - \left[ a \left( b + d \right) - b \left( a + c \right) \right] i}{{\left( a + c \right)}^{2} + {\left( b + d \right)}^{2}} \\ & = \frac {\left[ ( {a}^{2} + {b}^{2} ) + \left( a c + bd \right) \right] - \left( a d - b c \right) i}{( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} ) + 2 \left( a c + b d \right)}. \end{align}$$

For the sake of compactness, define $r = p / (p + q)$.

$\quad 1)$ Since $\mathfrak {Re} \left( r \right) < 0$, we know $( {a}^{2} + {b}^{2} ) + \left( a c + bd \right) < 0$.

$\quad 2)$ From $\mathfrak {Im} \left( r \right) = 0$, we derive $a d = b c$.

It is given in the question that $a, c \ge 0$.

• If $a, c > 0$, then $b d > 0$. Plug this back into $(1)$ and see that $(1)$ will be violated.

• If $a = 0$ and $c > 0$, then $b = 0$; however, still, $(1)$ will be violated, so this is impossible.

• The same thing will happen if $a > 0$ and $c = 0$.

• If $a = c = 0$, then $(2)$ will be satisfied. For $(1)$ to be true, we need ${b}^{2} + b d < 0$. If $b > 0$, then we need $b + d < 0$. If $b < 0$, then we need $b + d > 0$.