Find a complex number z given that 2+4i / z = 1+3i / 2.

The equation is:

$$z+4i = \frac{1+3i}{2\bar{z}}$$

Multiply both sides by $2\bar{z}$ to clear the denominator:

$$2\bar{z}(z+4i) = 1+3i$$ $$2z\bar{z} + 8i\bar{z} = 1+3i$$

Recall that $z\bar{z} = x^2+y^2$.

$$2(x^2+y^2) + 8i(x-yi) = 1+3i$$

Expand the left side:

$$2x^2+2y^2 + 8ix - 8i^2y = 1+3i$$

Since $i^2 = -1$:

$$2x^2+2y^2 + 8ix + 8y = 1+3i$$

For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.
Equating the real parts:

$$2x^2+2y^2+8y = 1 \quad \text{(Equation 1)}$$

Equating the imaginary parts:

$$8x = 3 \quad \text{(Equation 2)}$$

From Equation 2:

$$x = \frac{3}{8}$$

Substitute this value of $x$ into Equation 1:

$$2\left(\frac{3}{8}\right)^2 + 2y^2 + 8y = 1$$ $$2\left(\frac{9}{64}\right) + 2y^2 + 8y = 1$$ $$\frac{9}{32} + 2y^2 + 8y = 1$$

Rearrange the terms to form a quadratic equation in $y$:

$$2y^2 + 8y + \frac{9}{32} - 1 = 0$$ $$2y^2 + 8y - \frac{23}{32} = 0$$

Multiply the entire equation by 32 to eliminate the fraction:

$$64y^2 + 256y - 23 = 0$$

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=64$, $b=256$, $c=-23$:

$$y = \frac{-256 \pm \sqrt{256^2 - 4(64)(-23)}}{2(64)}$$ $$y = \frac{-256 \pm \sqrt{65536 + 5888}}{128}$$ $$y = \frac{-256 \pm \sqrt{71424}}{128}$$

Simplify the square root:

$$\sqrt{71424} = \sqrt{256 \times 279} = 16\sqrt{279}$$

Further simplify $\sqrt{279} = \sqrt{9 \times 31} = 3\sqrt{31}$.
So, $\sqrt{71424} = 16 \times 3\sqrt{31} = 48\sqrt{31}$.
Substitute this back into the expression for $y$:

$$y = \frac{-256 \pm 48\sqrt{31}}{128}$$

Divide the numerator and denominator by 16:

$$y = \frac{-16 \pm 3\sqrt{31}}{8}$$

We have $x = \frac{3}{8}$ and two possible values for $y$. Therefore, there are two possible complex numbers $z$:

$$z_1 = \frac{3}{8} + \left(\frac{-16 + 3\sqrt{31}}{8}\right)i$$ $$z_2 = \frac{3}{8} + \left(\frac{-16 - 3\sqrt{31}}{8}\right)i$$

The complex numbers are:

$$z = \frac{3}{8} + \left(\frac{-16 \pm 3\sqrt{31}}{8}\right)i$$