1. **Calculate the expression** $$\frac{7 - 5i}{129 + 6i} + \frac{1111 + 47i}{20031}$$ 2. **Find the square roots of** $$1 + i\sqrt{3}$$ in the form $$a + bi$$ where $$a,b \in \mathbb{R}$$. --- ### Step 1: Calculate $$\frac{7 - 5i}{129 + 6i}$$ Use the formula for division of complex numbers: $$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}$$ Here, $$a=7$$, $$b=-5$$, $$c=129$$, $$d=6$$. Calculate denominator: $$129^2 + 6^2 = 16641 + 36 = 16677$$ Calculate numerator: $$(7 - 5i)(129 - 6i) = 7 \times 129 - 7 \times 6i - 5i \times 129 + 5i \times 6i$$ $$= 903 - 42i - 645i + 30i^2$$ Recall $$i^2 = -1$$, so: $$= 903 - 42i - 645i - 30 = (903 - 30) + (-42 - 645)i = 873 - 687i$$ So: $$\frac{7 - 5i}{129 + 6i} = \frac{873 - 687i}{16677} = \frac{873}{16677} - \frac{687}{16677}i$$ Simplify fractions by dividing numerator and denominator by 3: $$\frac{873}{16677} = \frac{291}{5559}, \quad \frac{687}{16677} = \frac{229}{5559}$$ So: $$= \frac{291}{5559} - \frac{229}{5559}i$$ --- ### Step 2: Calculate $$\frac{1111 + 47i}{20031}$$ Divide real and imaginary parts by 20031: $$= \frac{1111}{20031} + \frac{47}{20031}i$$ --- ### Step 3: Add the two results: Real part: $$\frac{291}{5559} + \frac{1111}{20031}$$ Imaginary part: $$-\frac{229}{5559} + \frac{47}{20031}$$ Find common denominators: $$5559 \times 20031 = 111,393,729$$ Convert fractions: $$\frac{291}{5559} = \frac{291 \times 20031}{111,393,729} = \frac{5,829,021}{111,393,729}$$ $$\frac{1111}{20031} = \frac{1111 \times 5559}{111,393,729} = \frac{6,174,249}{111,393,729}$$ Sum real parts: $$\frac{5,829,021 + 6,174,249}{111,393,729} = \frac{12,003,270}{111,393,729}$$ Similarly for imaginary parts: $$-\frac{229}{5559} = -\frac{229 \times 20031}{111,393,729} = -\frac{4,587,699}{111,393,729}$$ $$\frac{47}{20031} = \frac{47 \times 5559}{111,393,729} = \frac{261,273}{111,393,729}$$ Sum imaginary parts: $$\frac{-4,587,699 + 261,273}{111,393,729} = \frac{-4,326,426}{111,393,729}$$ --- ### Step 4: Final answer for the first problem: $$\boxed{\frac{12,003,270}{111,393,729} - \frac{4,326,426}{111,393,729}i}$$ --- ### Step 5: Find square roots of $$1 + i\sqrt{3}$$ Let the root be $$z = a + bi$$, then: $$z^2 = (a + bi)^2 = a^2 + 2abi + b^2 i^2 = a^2 - b^2 + 2ab i$$ Equate real and imaginary parts: Real: $$a^2 - b^2 = 1$$ Imaginary: $$2ab = \sqrt{3}$$ From imaginary part: $$b = \frac{\sqrt{3}}{2a}$$ Substitute into real part: $$a^2 - \left(\frac{\sqrt{3}}{2a}\right)^2 = 1$$ $$a^2 - \frac{3}{4a^2} = 1$$ Multiply both sides by $$4a^2$$: $$4a^4 - 3 = 4a^2$$ Rearranged: $$4a^4 - 4a^2 - 3 = 0$$ Let $$x = a^2$$: $$4x^2 - 4x - 3 = 0$$ Solve quadratic: $$x = \frac{4 \pm \sqrt{16 + 48}}{8} = \frac{4 \pm \sqrt{64}}{8} = \frac{4 \pm 8}{8}$$ Two solutions: $$x_1 = \frac{12}{8} = \frac{3}{2}$$ $$x_2 = \frac{-4}{8} = -\frac{1}{2}$$ (discard negative since $$a^2 \geq 0$$) So: $$a^2 = \frac{3}{2} \Rightarrow a = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}$$ Find $$b$$: $$b = \frac{\sqrt{3}}{2a} = \frac{\sqrt{3}}{2 \times \pm \frac{\sqrt{6}}{2}} = \pm \frac{\sqrt{3}}{\sqrt{6}} = \pm \sqrt{\frac{3}{6}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ Check signs to satisfy imaginary part positive: $$2ab = \sqrt{3} > 0$$ If $$a > 0$$, then $$b > 0$$. Thus the two square roots are: $$\boxed{\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i \quad \text{and} \quad -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i}$$