Show from the power series (8.1) that 1 e^z_(1)cdot e^z_(2)=e^z_(1+z_(2)) 2 (d)/(dz)e^z=e^z

Let's break down how to demonstrate these properties of the exponential function using its power series representation.**1. e^(z₁)*e^(z₂) = e^(z₁+z₂) ****Power Series Representation:**The power series representation of the exponential function is:```e^z = 1 + z + (z^2)/2! + (z^3)/3! + ... = Σ (z^n)/n! ```where the summation (Σ) goes from n = 0 to infinity.**Proof:**1. **Expand e^(z₁) and e^(z₂):** ``` e^(z₁) = 1 + z₁ + (z₁^2)/2! + (z₁^3)/3! + ... e^(z₂) = 1 + z₂ + (z₂^2)/2! + (z₂^3)/3! + ... ```2. **Multiply the series:** To multiply these infinite series, we need to consider all possible combinations of terms. This is a bit tedious to write out fully, but the pattern becomes clear: ``` e^(z₁) * e^(z₂) = (1 + z₁ + (z₁^2)/2! + ...) * (1 + z₂ + (z₂^2)/2! + ...) ``` Expanding this, we get terms like: * 1 * 1 * 1 * z₂ * z₁ * 1 * z₁ * z₂ * (z₁^2)/2! * 1 * (z₁^2)/2! * z₂ * ... and so on.3. **Collect terms with the same power of (z₁ + z₂):** Notice that terms like z₁ * z₂ and z₂ * z₁ both contribute to the coefficient of (z₁ + z₂)^2. Similarly, terms like (z₁^2)/2! * 1, (z₁^2)/2! * z₂, z₁ * (z₂^2)/2!, and (z₂^2)/2! * 1 all contribute to the coefficient of (z₁ + z₂)^3. In general, the coefficient of (z₁ + z₂)^n will be the sum of all terms where the powers of z₁ and z₂ add up to n. This is exactly the same as the coefficient of z^n in the expansion of e^(z₁ + z₂).4. **Conclusion:** Therefore, the product of the power series for e^(z₁) and e^(z₂) is identical to the power series for e^(z₁ + z₂). This proves the identity: ``` e^(z₁) * e^(z₂) = e^(z₁ + z₂) ```**2. d/dz e^z = e^z****Proof:**1. **Differentiate the power series term by term:** ``` d/dz (e^z) = d/dz (1 + z + (z^2)/2! + (z^3)/3! + ...) ``` Since differentiation is a linear operation, we can differentiate each term individually: ``` d/dz (e^z) = 0 + 1 + 2z/2! + 3z^2/3! + ... ```2. **Simplify:** Notice that the derivative of (z^n)/n! is (n*z^(n-1))/n! = z^(n-1)/(n-1)!. This means the derivative of the power series is: ``` d/dz (e^z) = 1 + z + (z^2)/2! + (z^3)/3! + ... ```3. **Recognize the result:** The result is exactly the same as the original power series for e^z. Therefore: ``` d/dz (e^z) = e^z ```**Key Points:*** The power series representation of the exponential function is a powerful tool for proving its properties.* The ability to differentiate and multiply power series term by term is essential for these proofs.* These properties are fundamental to the understanding and application of the exponential function in mathematics, physics, and other fields.