We are given the equation $\frac{1}{x+iy} + \frac{1}{1+3i} = 1$ and we need to solve for $x$ and $y$.

1. Problem Description

We are given the equation $\frac{1}{x+iy} + \frac{1}{1+3i} = 1$ and we need to solve for $x$ and $y$.

2. Solution Steps

First, isolate the term with $x$ and $y$:

$\frac{1}{x+iy} = 1 - \frac{1}{1+3i}$

$\frac{1}{x+iy} = \frac{1+3i - 1}{1+3i}$

$\frac{1}{x+iy} = \frac{3i}{1+3i}$

Next, take the reciprocal of both sides:

$x+iy = \frac{1+3i}{3i}$

$x+iy = \frac{1}{3i} + \frac{3i}{3i}$

$x+iy = \frac{1}{3i} + 1$

To simplify the complex number $\frac{1}{3i}$, multiply the numerator and denominator by the complex conjugate of the denominator, which is $-3i$:

$\frac{1}{3i} = \frac{1}{3i} \cdot \frac{-3i}{-3i} = \frac{-3i}{-9i^2} = \frac{-3i}{9} = -\frac{i}{3}$

Substitute this back into the equation:

$x+iy = -\frac{i}{3} + 1$

$x+iy = 1 - \frac{1}{3}i$

Now, equate the real and imaginary parts:

$x = 1$

$y = -\frac{1}{3}$

3. Final Answer

$x = 1$ and $y = -\frac{1}{3}$