Iqic: Division 1. Simplify: a. \( \frac{z-3 i}{i} \) b) \( \frac{1-7 i}{-2-4 i} \) c \( \frac{1-2 i}{i-5} \) d. \( \frac{\sqrt{2}+5 i}{-3 i+\sqrt{2}} \)

Sure, let's simplify each of the given complex fractions step by step. --- ### **Problem 1. Simplify:** #### **a. \( \frac{z - 3i}{i} \)** **Solution:** To simplify \( \frac{z - 3i}{i} \), we can separate the terms in the numerator: \[ \frac{z - 3i}{i} = \frac{z}{i} - \frac{3i}{i} \] Simplifying each term: 1. \( \frac{z}{i} \): Remember that \( \frac{1}{i} = -i \) (since \( i \times (-i) = 1 \)). \[ \frac{z}{i} = z \times \frac{1}{i} = z \times (-i) = -iz \] 2. \( \frac{3i}{i} \): \[ \frac{3i}{i} = 3 \times \frac{i}{i} = 3 \times 1 = 3 \] Putting it all together: \[ \frac{z - 3i}{i} = -iz - 3 \] **Final Answer:** \[ \frac{z - 3i}{i} = -iz - 3 \] --- #### **b. \( \frac{1 - 7i}{-2 - 4i} \)** **Solution:** To simplify the complex fraction, multiply the numerator and the denominator by the **conjugate** of the denominator. 1. **Conjugate of the denominator** \( -2 - 4i \) is \( -2 + 4i \). 2. Multiply numerator and denominator by \( -2 + 4i \): \[ \frac{1 - 7i}{-2 - 4i} \times \frac{-2 + 4i}{-2 + 4i} = \frac{(1 - 7i)(-2 + 4i)}{(-2 - 4i)(-2 + 4i)} \] 3. **Expand the numerator:** \[ (1)(-2) + (1)(4i) + (-7i)(-2) + (-7i)(4i) = -2 + 4i + 14i - 28i^2 \] Since \( i^2 = -1 \): \[ -2 + 4i + 14i - 28(-1) = -2 + 18i + 28 = 26 + 18i \] 4. **Expand the denominator:** \[ (-2)^2 - (4i)^2 = 4 - 16i^2 = 4 - 16(-1) = 4 + 16 = 20 \] 5. **Combine the results:** \[ \frac{26 + 18i}{20} = \frac{26}{20} + \frac{18}{20}i = \frac{13}{10} + \frac{9}{10}i \] **Final Answer:** \[ \frac{1 - 7i}{-2 - 4i} = \frac{13}{10} + \frac{9}{10}i \] --- #### **c. \( \frac{1 - 2i}{i - 5} \)** **Solution:** Again, we'll multiply the numerator and the denominator by the **conjugate** of the denominator. 1. **Conjugate of the denominator** \( i - 5 \) is \( -5 - i \). 2. Multiply numerator and denominator by \( -5 - i \): \[ \frac{1 - 2i}{i - 5} \times \frac{-5 - i}{-5 - i} = \frac{(1 - 2i)(-5 - i)}{(i - 5)(-5 - i)} \] 3. **Expand the numerator:** \[ (1)(-5) + (1)(-i) + (-2i)(-5) + (-2i)(-i) = -5 - i + 10i + 2i^2 \] Since \( i^2 = -1 \): \[ -5 - i + 10i + 2(-1) = -5 + 9i - 2 = -7 + 9i \] 4. **Expand the denominator:** \[ (i)(-5) + (i)(-i) + (-5)(-5) + (-5)(-i) = -5i -i^2 + 25 + 5i \] Again, \( i^2 = -1 \): \[ -5i + 5i + 1 + 25 = 26 \] 5. **Combine the results:** \[ \frac{-7 + 9i}{26} = -\frac{7}{26} + \frac{9}{26}i \] **Final Answer:** \[ \frac{1 - 2i}{i - 5} = -\frac{7}{26} + \frac{9}{26}i \] --- #### **d. \( \frac{\sqrt{2} + 5i}{-3i + \sqrt{2}} \)** **Solution:** We'll multiply the numerator and the denominator by the **conjugate** of the denominator. 1. **Conjugate of the denominator** \( -3i + \sqrt{2} \) is \( \sqrt{2} + 3i \). 2. Multiply numerator and denominator by \( \sqrt{2} + 3i \): \[ \frac{\sqrt{2} + 5i}{-3i + \sqrt{2}} \times \frac{\sqrt{2} + 3i}{\sqrt{2} + 3i} = \frac{(\sqrt{2} + 5i)(\sqrt{2} + 3i)}{(-3i + \sqrt{2})(\sqrt{2} + 3i)} \] 3. **Expand the numerator:** \[ (\sqrt{2})(\sqrt{2}) + (\sqrt{2})(3i) + (5i)(\sqrt{2}) + (5i)(3i) = 2 + 3\sqrt{2}i + 5\sqrt{2}i + 15i^2 \] Since \( i^2 = -1 \): \[ 2 + 8\sqrt{2}i + 15(-1) = 2 + 8\sqrt{2}i - 15 = -13 + 8\sqrt{2}i \] 4. **Expand the denominator:** \[ (-3i)(\sqrt{2}) + (-3i)(3i) + (\sqrt{2})(\sqrt{2}) + (\sqrt{2})(3i) = -3\sqrt{2}i - 9i^2 + 2 + 3\sqrt{2}i \] Again, \( i^2 = -1 \): \[ -3\sqrt{2}i + 3\sqrt{2}i + 9 + 2 = 11 \] 5. **Combine the results:** \[ \frac{-13 + 8\sqrt{2}i}{11} = -\frac{13}{11} + \frac{8\sqrt{2}}{11}i \] **Final Answer:** \[ \frac{\sqrt{2} + 5i}{-3i + \sqrt{2}} = -\frac{13}{11} + \frac{8\sqrt{2}}{11}i \] --- ### **Summary of All Simplified Forms:** - **a.** \[ \frac{z - 3i}{i} = -iz - 3 \] - **b.** \[ \frac{1 - 7i}{-2 - 4i} = \frac{13}{10} + \frac{9}{10}i \] - **c.** \[ \frac{1 - 2i}{i - 5} = -\frac{7}{26} + \frac{9}{26}i \] - **d.** \[ \frac{\sqrt{2} + 5i}{-3i + \sqrt{2}} = -\frac{13}{11} + \frac{8\sqrt{2}}{11}i \]