Solve for z: (3-2i)/(4i+2) = (4−i)/z

The given equation is:

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

Rearrange the equation to solve for $\bar{z}$:

$$\bar{z} = \frac{4-i}{\frac{3-2i}{4i+2}}$$ $$\bar{z} = (4-i) \times \frac{4i+2}{3-2i}$$ $$\bar{z} = \frac{(4-i)(2+4i)}{3-2i}$$

$$(4-i)(2+4i) = 4(2) + 4(4i) - i(2) - i(4i)$$ $$= 8 + 16i - 2i - 4i^2$$

Since $i^2 = -1$:

$$= 8 + 14i - 4(-1)$$ $$= 8 + 14i + 4$$ $$= 12 + 14i$$

So, the equation becomes:

$$\bar{z} = \frac{12+14i}{3-2i}$$

Multiply the numerator and denominator by the conjugate of the denominator, which is $3+2i$.

$$\bar{z} = \frac{12+14i}{3-2i} \times \frac{3+2i}{3+2i}$$

Numerator:

$$(12+14i)(3+2i) = 12(3) + 12(2i) + 14i(3) + 14i(2i)$$ $$= 36 + 24i + 42i + 28i^2$$ $$= 36 + 66i + 28(-1)$$ $$= 36 + 66i - 28$$ $$= 8 + 66i$$

Denominator:

$$(3-2i)(3+2i) = 3^2 - (2i)^2$$ $$= 9 - 4i^2$$ $$= 9 - 4(-1)$$ $$= 9 + 4$$ $$= 13$$

So, we have:

$$\bar{z} = \frac{8+66i}{13}$$ $$\bar{z} = \frac{8}{13} + \frac{66}{13}i$$

If $\bar{z} = x+yi$, then $z = x-yi$.

$$z = \frac{8}{13} - \frac{66}{13}i$$