Let's solve each part step-by-step as requested. --- ## **4. Assume that \( s = \sum_{n=0}^N x^n \).** ### (a) \( x = 1 + j \), \( N = 10 \) #### **Step 1: Write the sum in closed form (geometric series)** The sum is a geometric series: \[ s = \sum_{n=0}^N x^n = \frac{x^{N+1} - 1}{x - 1} \] **Explanation:** This is the formula for the sum of a geometric series with ratio \( x \) and \( N+1 \) terms. --- #### **Step 2: Substitute the given values** Substitute \( x = 1 + j \), \( N = 10 \): \[ s = \frac{(1 + j)^{11} - 1}{(1 + j) - 1} = \frac{(1 + j)^{11} - 1}{j} \] **Explanation:** Replace \( x \) and \( N \) in the formula, and note \( x - 1 = j \). --- **Final answer for (a):** \[ \boxed{s = \frac{(1 + j)^{11} - 1}{j}} \] --- ### (b) \( x = 0.5 - j \), \( N = 7 \) #### **Step 1: Write the sum in closed form** Again, use the geometric series formula: \[ s = \frac{x^{N+1} - 1}{x - 1} \] **Explanation:** Same formula, just with new values of \( x \) and \( N \). --- #### **Step 2: Substitute the given values** Substitute \( x = 0.5 - j \), \( N = 7 \): \[ s = \frac{(0.5 - j)^{8} - 1}{(0.5 - j) - 1} = \frac{(0.5 - j)^{8} - 1}{-0.5 - j} \] **Explanation:** Replace \( x \) and \( N \), and simplify \( x - 1 = 0.5 - j - 1 = -0.5 - j \). --- **Final answer for (b):** \[ \boxed{s = \frac{(0.5 - j)^8 - 1}{-0.5 - j}} \] --- ## **5. Using Euler's formula, show that \( e^{j2k\pi} = 1 \).** #### **Step 1: Recall Euler's formula** Euler's formula: \( e^{j\theta} = \cos\theta + j\sin\theta \). **Explanation:** This relates complex exponentials to trigonometric functions. --- #### **Step 2: Apply \( \theta = 2k\pi \)** \[ e^{j2k\pi} = \cos(2k\pi) + j\sin(2k\pi) = 1 + j \cdot 0 = 1 \] **Explanation:** \(\cos(2k\pi) = 1\), \(\sin(2k\pi) = 0\) for any integer \( k \). --- **Final answer:** \[ \boxed{e^{j2k\pi} = 1} \] --- ## **6. Using Euler's formula, show that \( e^{jk\pi} = (-1)^k \).** #### **Step 1: Apply Euler's formula** \[ e^{jk\pi} = \cos(k\pi) + j\sin(k\pi) \] **Explanation:** Direct substitution into Euler's formula. --- #### **Step 2: Evaluate for integer \( k \)** \[ \cos(k\pi) = (-1)^k, \quad \sin(k\pi) = 0 \] So, \( e^{jk\pi} = (-1)^k \). **Explanation:** \(\cos(k\pi)\) alternates between \( 1 \) and \( -1 \) as \( k \) changes, while \(\sin(k\pi) = 0 \) for integer \( k \). --- **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: Express \( 1+j \) in polar form** \[ 1+j = \sqrt{2}e^{j\frac{\pi}{4}} \] So, \[ (1+j)^k = (\sqrt{2})^k \cdot e^{jk\frac{\pi}{4}} \] **Explanation:** Convert to polar: modulus is \( \sqrt{2} \), argument is \( \frac{\pi}{4} \). --- #### **Step 2: Set argument to \( n\pi \) for real result** For \( (1+j)^k \) to be real, the exponent must be \( e^{j n\pi} \), i.e., \( k\frac{\pi}{4} = n\pi \) for integer \( n \). \[ k = 4n \] **Explanation:** This ensures the imaginary part is zero and the result is real. --- **Final answer:** \[ \boxed{k = 4n, \text{ where } n \in \mathbb{Z}} \] or more explicitly, **all integer multiples of 4.**

# Step-by-step Solutions --- ## 4. Sum \( s = \sum_{n=0}^N x^n \) ### (a) \( x = 1 + j \), \( N=10 \) ### Step 1: Write in closed form \[ s = \frac{x^{N+1} - 1}{x - 1} \] **Explanation:** Sum of a geometric series with ratio \( x \) over \( N+1 \) terms. ### Step 2: Substitute values \[ s = \frac{(1 + j)^{11} - 1}{(1 + j) - 1} = \frac{(1 + j)^{11} - 1}{j} \] **Final answer:** \[ \boxed{s = \frac{(1 + j)^{11} - 1}{j}} \] --- ### (b) \( x=0.5 - j \), \( N=7 \) ### Step 1: Write in closed form \[ s = \frac{x^{8} - 1}{x - 1} \] ### Step 2: Substitute values \[ s = \frac{(0.5 - j)^{8} - 1}{(0.5 - j) - 1} = \frac{(0.5 - j)^{8} - 1}{-0.5 - j} \] **Final answer:** \[ \boxed{s = \frac{(0.5 - j)^{8} - 1}{-0.5 - j}} \] --- ## 5. Show \( e^{j2k\pi} = 1 \) using Euler's formula ### Step 1: Euler's formula states: \[ e^{j\theta} = \cos\theta + j\sin\theta \] ### Step 2: Set \( \theta = 2k\pi \): \[ e^{j2k\pi} = \cos(2k\pi) + j\sin(2k\pi) = 1 + j \cdot 0 = 1 \] **Final answer:** \[ \boxed{e^{j2k\pi} = 1} \] --- ## 6. Show \( e^{jk\pi} = (-1)^k \) using Euler's formula ### Step 1: \[ e^{jk\pi} = \cos(k\pi) + j\sin(k\pi) \] ### Step 2: Since \( \sin(k\pi) = 0 \), and \( \cos(k\pi) = (-1)^k \), \[ e^{jk\pi} = (-1)^k \] **Final answer:** \[ \boxed{e^{jk\pi} = (-1)^k} \] --- ## 7. Find all \( k \) such that \( (1 + j)^k \) is real ### Step 1: Express \( 1 + j \) in polar form: \[ 1 + j = \sqrt{2} e^{j\pi/4} \] Then: \[ (1 + j)^k = (\sqrt{2})^k e^{jk\pi/4} \] which simplifies to: \[ 2^{k/2} e^{jk\pi/4} \] ### Step 2: For the entire expression to be real, the exponential part must be real, i.e., \[ e^{jk\pi/4} = \pm 1 \] This occurs when: \[ jk\frac{\pi}{4} = n\pi \quad \text{for } n \in \mathbb{Z} \] \[ \Rightarrow k \frac{\pi}{4} = n \pi \quad \Rightarrow \quad k = 4n \] **Final answer:** \[ \boxed{k = 4n, \quad n \in \mathbb{Z}} \] --- **Summary:** - **(a)** \( s = \frac{(1 + j)^{11} - 1}{j} \) - **(b)** \( s = \frac{(0.5 - j)^{8} - 1}{-0.5 - j} \) - **Euler's formulas:** - \( e^{j2k\pi} = 1 \) - \( e^{jk\pi} = (-1)^k \) - **Values of \( k \) for real \( (1+j)^k \):** - \( k = 4n \), \( n \in \mathbb{Z} \)

## Final Answers 1. **Sum \( s \):** - (a) \( s = \frac{(1 + j)^{11} - 1}{j} \) - (b) \( s = \frac{(0.5 - j)^{8} - 1}{-0.5 - j} \) 2. **Using Euler's formula:** - \( e^{j2k\pi} = 1 \) - \( e^{jk\pi} = (-1)^k \) 3. **Values of \( k \) for which \( (1 + j)^k \) is real:** \[ \boxed{k = 4n, \quad n \in \mathbb{Z}} \]