1. **Problem Statement:** Given the function $f(z) = e^x (\cos x + i \sin y)$ where $z = x + iy$, show that: (a) $f(z_1 + z_2) = f(z_1) f(z_2)$ (b) $\overline{f(z)} = f(\overline{z})$ 2. **Recall:** - For complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, addition is component-wise: $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$. - The conjugate of $z = x + iy$ is $\overline{z} = x - iy$. - The conjugate of a product is the product of conjugates: $\overline{ab} = \overline{a} \cdot \overline{b}$. 3. **Part (a): Show $f(z_1 + z_2) = f(z_1) f(z_2)$** - Write $z_1 = x_1 + iy_1$, $z_2 = x_2 + iy_2$. - Then $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$. - Compute $f(z_1 + z_2)$: $$f(z_1 + z_2) = e^{x_1 + x_2} \left( \cos(x_1 + x_2) + i \sin(y_1 + y_2) \right)$$ - Compute $f(z_1) f(z_2)$: $$f(z_1) f(z_2) = e^{x_1} (\cos x_1 + i \sin y_1) \times e^{x_2} (\cos x_2 + i \sin y_2) = e^{x_1 + x_2} (\cos x_1 + i \sin y_1)(\cos x_2 + i \sin y_2)$$ - Multiply the complex terms: $$(\cos x_1 + i \sin y_1)(\cos x_2 + i \sin y_2) = \cos x_1 \cos x_2 + i \cos x_1 \sin y_2 + i \sin y_1 \cos x_2 + i^2 \sin y_1 \sin y_2$$ - Since $i^2 = -1$, this becomes: $$\cos x_1 \cos x_2 - \sin y_1 \sin y_2 + i (\cos x_1 \sin y_2 + \sin y_1 \cos x_2)$$ - Note that $\cos(x_1 + x_2) = \cos x_1 \cos x_2 - \sin x_1 \sin x_2$, but here the sine arguments differ ($y$ vs $x$), so the equality does not hold in general. **Conclusion:** The given function $f(z)$ does not satisfy $f(z_1 + z_2) = f(z_1) f(z_2)$ unless $x = y$ or under special conditions. So part (a) is false in general. 4. **Part (b): Show $\overline{f(z)} = f(\overline{z})$** - Compute $\overline{f(z)}$: $$\overline{f(z)} = \overline{e^x (\cos x + i \sin y)} = e^x (\cos x - i \sin y)$$ - Compute $f(\overline{z})$: $$\overline{z} = x - iy$$ $$f(\overline{z}) = e^x (\cos x + i \sin(-y)) = e^x (\cos x - i \sin y)$$ - Since $\sin(-y) = -\sin y$, we have: $$\overline{f(z)} = f(\overline{z})$$ **Final answers:** (a) $f(z_1 + z_2) \neq f(z_1) f(z_2)$ in general. (b) $\overline{f(z)} = f(\overline{z})$ holds true.