1. Problem statement: Given the complex number $z_1 = 2 + i$, find complex numbers $z = x + iy$ satisfying the given conditions. 2. Recall that for complex numbers $z = x + iy$ and $w = u + iv$, the product and quotient are: $$z \cdot w = (xu - yv) + i(xv + yu)$$ $$\frac{z}{w} = \frac{z \cdot \overline{w}}{|w|^2} = \frac{(x + iy)(u - iv)}{u^2 + v^2} = \frac{(xu + yv) + i(yu - xv)}{u^2 + v^2}$$ where $\overline{w} = u - iv$ is the conjugate of $w$. 3. For $z_1 = 2 + i$, we have $u=2$, $v=1$, and $|z_1|^2 = 2^2 + 1^2 = 5$. 4. a) Find $z$ such that $$\text{Re}\left(\frac{z}{z_1}\right) = -5 \quad \text{and} \quad \text{Im}\left(\frac{z}{z_1}\right) = 3.$$ Using the formula for division: $$\frac{z}{z_1} = \frac{(2x + y) + i(y \cdot 2 - x \cdot 1)}{5} = \frac{(2x + y) + i(2y - x)}{5}.$$ So, $$\text{Re}\left(\frac{z}{z_1}\right) = \frac{2x + y}{5} = -5 \implies 2x + y = -25,$$ $$\text{Im}\left(\frac{z}{z_1}\right) = \frac{2y - x}{5} = 3 \implies 2y - x = 15.$$ Solve the system: From second equation: $x = 2y - 15$. Substitute into first: $$2(2y - 15) + y = -25 \implies 4y - 30 + y = -25 \implies 5y = 5 \implies y = 1.$$ Then, $$x = 2(1) - 15 = 2 - 15 = -13.$$ 5. b) Find $z$ such that $$\text{Re}(z \cdot z_1) = 1 \quad \text{and} \quad \text{Im}\left(\frac{z}{z_1}\right) = -5.$$ Calculate $z \cdot z_1$: $$z \cdot z_1 = (x + iy)(2 + i) = (2x - y) + i(x + 2y).$$ So, $$\text{Re}(z \cdot z_1) = 2x - y = 1,$$ and from division formula above, $$\text{Im}\left(\frac{z}{z_1}\right) = \frac{2y - x}{5} = -5 \implies 2y - x = -25.$$ Solve the system: From second equation: $x = 2y + 25$. Substitute into first: $$2(2y + 25) - y = 1 \implies 4y + 50 - y = 1 \implies 3y = -49 \implies y = -\frac{49}{3}.$$ Then, $$x = 2\left(-\frac{49}{3}\right) + 25 = -\frac{98}{3} + 25 = -\frac{98}{3} + \frac{75}{3} = -\frac{23}{3}.$$ Final answers: - a) $z = -13 + i$. - b) $z = -\frac{23}{3} - i\frac{49}{3}$.