Simplify the equation (x+iy)/(x-iy) + (x-iy)/(x+iy) +…

Simplify the equation $$ \frac{x+iy}{x-iy} + \frac{x-iy}{x+iy} + 4a\left(\frac{1}{x+iy}+\frac{1}{x-iy}\right) = 2. $$

Step 1 · Rewrite with $z=x+iy$

With $\bar z=x-iy$, the equation becomes:

$$ \frac{z}{\bar z}+\frac{\bar z}{z} + 4a\left(\frac1z+\frac1{\bar z}\right) = 2 $$

Step 2 · Simplify $\dfrac{z}{\bar z}+\dfrac{\bar z}{z}$

$$ \frac{z}{\bar z}+\frac{\bar z}{z} = \frac{z^2+\bar z^2}{z\bar z} $$

Since $z\bar z=|z|^2=x^2+y^2$, and

$$ z^2=(x^2-y^2)+2ixy, \qquad \bar z^2=(x^2-y^2)-2ixy \quad\Longrightarrow\quad z^2+\bar z^2=2(x^2-y^2) $$

we get:

$$ \frac{z}{\bar z}+\frac{\bar z}{z} = \frac{2(x^2-y^2)}{x^2+y^2} $$

Step 3 · Simplify $\dfrac1z+\dfrac1{\bar z}$

$$ \frac1z+\frac1{\bar z} = \frac{z+\bar z}{z\bar z} = \frac{2x}{x^2+y^2} $$

since $z+\bar z = (x+iy)+(x-iy)=2x$.

Step 4 · Substitute back

$$ \frac{2(x^2-y^2)}{x^2+y^2} + 4a\cdot\frac{2x}{x^2+y^2} = 2 \quad\Longrightarrow\quad \frac{2(x^2-y^2)+8ax}{x^2+y^2} = 2 $$

Step 5 · Clear the denominator

$$ 2(x^2-y^2)+8ax = 2(x^2+y^2) $$

Step 6 · Expand and cancel

$$ 2x^2-2y^2+8ax = 2x^2+2y^2 \quad\Longrightarrow\quad 8ax = 4y^2 $$

Step 7 · Final simplified form

$$ \boxed{y^2 = 2ax} \qquad\text{equivalently}\qquad \boxed{y^2-2ax=0} $$

Step 8 · Geometric interpretation

For $a\neq0$, $y^2=2ax$ is a parabola with vertex at the origin, axis of symmetry along the $x$-axis, and semi-latus rectum parameter $p=\frac{a}{2}$ (comparing to $(y-k)^2=4p(x-h)$). It opens to the right if $a>0$ and to the left if $a<0$.

Step 9 · Numerical check

Take $x=2a$, $y=2a$ (with $a>0$): then $y^2=4a^2=2a(2a)=2ax$ ✓. Substituting into the original equation:

$$ \frac{2(x^2-y^2)}{x^2+y^2}+\frac{8ax}{x^2+y^2} = \frac{2(4a^2-4a^2)}{8a^2}+\frac{16a^2}{8a^2} = 0+2 = 2 \checkmark $$

Step 10 · Special cases

For $a=0$: the equation $\frac{2(x^2-y^2)}{x^2+y^2}=2$ gives $x^2-y^2=x^2+y^2$, i.e. $y=0$.