Question: User: $\frac{x+yi}{1+5i} = \frac{3-2i}{2}$
$\frac{x+yi}{1+5i} \leftrightarrow \frac{3+2i}{-i}$
$(x+yi)(-i) = (1+5i)(3+2i)$
$-xi - yi^2 = 3 + 2i + 15i + 10i^2 \Rightarrow y - xi = -7 + 17i$
$y = -7, -x = 17$
$x = -17$
Ex1: If $\frac{2x-yi}{1+2i}$, $\frac{3-i}{1-i}$ are conjugated find the
real value of $x,y$
$\frac{2x-yi}{1+2i} = \frac{3-i}{1-i}$
$2x-yi \leftrightarrow 3+i$
$\frac{}{1+2i} \quad \frac{}{1+i}$ The question in red
1. **State the problem:**
Find the real values of $x$ and $y$ such that the complex fractions $\frac{2x - yi}{1 + 2i}$ and $\frac{3 - i}{1 - i}$ are conjugates.
2. **Recall the conjugate property:**
Two complex numbers $z$ and $w$ are conjugates if $w = \overline{z}$.
3. **Rewrite the problem:**
Given $\frac{2x - yi}{1 + 2i}$ and $\frac{3 - i}{1 - i}$ are conjugates, then:
$$\frac{2x - yi}{1 + 2i} = \overline{\frac{3 - i}{1 - i}} = \frac{3 + i}{1 + i}$$
4. **Simplify the right side:**
Multiply numerator and denominator by the conjugate of denominator:
$$\frac{3 + i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(3 + i)(1 - i)}{(1 + i)(1 - i)}$$
Calculate numerator:
$$(3 + i)(1 - i) = 3 \times 1 - 3i + i \times 1 - i^2 = 3 - 3i + i + 1 = 4 - 2i$$
Calculate denominator:
$$(1 + i)(1 - i) = 1 - i + i - i^2 = 1 + 1 = 2$$
So,
$$\frac{3 + i}{1 + i} = \frac{4 - 2i}{2} = 2 - i$$
5. **Set the equality:**
$$\frac{2x - yi}{1 + 2i} = 2 - i$$
6. **Multiply both sides by $1 + 2i$ to clear denominator:**
$$(2x - yi) = (2 - i)(1 + 2i)$$
Calculate right side:
$$(2 - i)(1 + 2i) = 2 \times 1 + 2 \times 2i - i \times 1 - i \times 2i = 2 + 4i - i - 2i^2 = 2 + 3i + 2 = 4 + 3i$$
7. **Equate real and imaginary parts:**
Left side: $2x - yi = 2x + (-y)i$
Right side: $4 + 3i$
So,
Real parts: $2x = 4 \Rightarrow x = 2$
Imaginary parts: $-y = 3 \Rightarrow y = -3$
**Final answer:**
$$x = 2, \quad y = -3$$