Let's solve each part step by step as requested: --- ### **4. Assume that \( s = \sum_{n=0}^{N} x^n \). Find \( s \) if:** #### **(a) \( x = 1 + j, N = 10 \)** **Step 1:** Recognize the formula for the sum of a geometric series: \[ s = \sum_{n=0}^{N} x^n = \frac{1-x^{N+1}}{1-x} \] **Explanation:** This is the standard formula for the sum of a geometric series where the first term is 1 and the ratio is \( x \). --- **Step 2:** Substitute the given values \( x = 1 + j \), \( N = 10 \): \[ s = \frac{1-(1+j)^{11}}{1-(1+j)} = \frac{1-(1+j)^{11}}{-j} \] **Explanation:** Plug in \( x \) and \( N+1 = 11 \). Simplifying the denominator: \( 1-(1+j) = -j \). **Final Answer (a):** \[ \boxed{s = \frac{1-(1+j)^{11}}{-j}} \] --- #### **(b) \( x = 0.5 - j, N = 7 \)** **Step 1:** Use the same geometric series formula: \[ s = \frac{1-x^{N+1}}{1-x} \] **Explanation:** The formula remains the same, but now with different values for \( x \) and \( N \). --- **Step 2:** Substitute \( x = 0.5 - j \), \( N = 7 \): \[ s = \frac{1-(0.5-j)^8}{1-(0.5-j)} \] \[ = \frac{1-(0.5-j)^8}{0.5+j} \] **Explanation:** Plug in \( x \) and \( N+1 = 8 \). Simplify \( 1-(0.5-j) = 0.5 + j \). **Final Answer (b):** \[ \boxed{s = \frac{1-(0.5-j)^8}{0.5+j}} \] --- ### **5. Using Euler's formula show that \( e^{j2k\pi} = 1 \).** **Step 1:** Recall Euler's formula: \( e^{j\theta} = \cos\theta + j\sin\theta \). **Explanation:** Euler's formula relates exponentials with imaginary arguments to trigonometric functions. --- **Step 2:** Substitute \( \theta = 2k\pi \): \[ e^{j2k\pi} = \cos(2k\pi) + j\sin(2k\pi) \] For any integer \( k \), \( \cos(2k\pi) = 1 \) and \( \sin(2k\pi) = 0 \): \[ e^{j2k\pi} = 1 + j \cdot 0 = 1 \] **Explanation:** Sine of any integer multiple of \( 2\pi \) is 0, cosine is 1. **Final Answer:** \[ \boxed{e^{j2k\pi} = 1} \] --- ### **6. Using Euler's formula show that \( e^{jk\pi} = (-1)^k \).** **Step 1:** Using Euler's formula: \( e^{j\theta} = \cos\theta + j\sin\theta \). **Explanation:** Apply Euler's formula to \( \theta = k\pi \). --- **Step 2:** Substitute \( \theta = k\pi \): \[ e^{jk\pi} = \cos(k\pi) + j\sin(k\pi) \] For integer \( k \), \( \cos(k\pi) = (-1)^k \) and \( \sin(k\pi) = 0 \): \[ e^{jk\pi} = (-1)^k + j \cdot 0 = (-1)^k \] **Explanation:** Cosine of \( k\pi \) alternates between 1 and -1 depending on \( k \). **Final Answer:** \[ \boxed{e^{jk\pi} = (-1)^k} \] --- ### **7. Find all values of \( k \) so that \( (1 + j)^k \) is a real number.** **Step 1:** Write \( 1 + j \) in polar form: \[ 1 + j = \sqrt{2} e^{j\frac{\pi}{4}} \] So, \[ (1+j)^k = \left(\sqrt{2}\right)^k e^{j k \frac{\pi}{4}} \] **Explanation:** Expressing in polar form makes exponentiation easier using De Moivre's theorem. --- **Step 2:** For the result to be real, the imaginary part must be zero, i.e., \( e^{j k \frac{\pi}{4}} \) must be real. This happens when: \[ k \frac{\pi}{4} = n\pi, \quad n \in \mathbb{Z} \] \[ k = 4n \] **Explanation:** The exponential is real when the angle is a multiple of \( \pi \). **Final Answer:** \[ \boxed{k = 4n \text{ where } n \text{ is any integer}} \]