1. **Problem Statement:** Find the multiplicative inverse $z^{-1}$ of the complex number $z = \sqrt{3} + i \sqrt{7}$ and prove that $zz^{-1} = 1$. 2. **Formula and Rules:** The multiplicative inverse of a complex number $z = a + bi$ is given by $$ z^{-1} = \frac{1}{z} = \frac{\overline{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2} $$ where $\overline{z}$ is the complex conjugate of $z$ and $|z|$ is the modulus. 3. **Calculate modulus and conjugate:** - $a = \sqrt{3}$, $b = \sqrt{7}$ - $|z|^2 = (\sqrt{3})^2 + (\sqrt{7})^2 = 3 + 7 = 10$ - $\overline{z} = \sqrt{3} - i \sqrt{7}$ 4. **Find the inverse:** $$ z^{-1} = \frac{\sqrt{3} - i \sqrt{7}}{10} = \frac{\sqrt{3}}{10} - i \frac{\sqrt{7}}{10} $$ 5. **Verify $zz^{-1} = 1$:** $$ zz^{-1} = \left(\sqrt{3} + i \sqrt{7}\right) \left(\frac{\sqrt{3}}{10} - i \frac{\sqrt{7}}{10}\right) = \frac{1}{10} \left(3 - i \sqrt{21} + i \sqrt{21} + 7\right) = \frac{1}{10} (3 + 7) = 1 $$ The imaginary parts cancel out, confirming the product is 1. **Final answer:** $$ z^{-1} = \frac{\sqrt{3}}{10} - i \frac{\sqrt{7}}{10} $$ and $$ zz^{-1} = 1 $$