1. The problem is to evaluate the expression $\sqrt{4 + 3i}$. 2. To find the square root of a complex number, we assume it can be written as $a + bi$ where $a$ and $b$ are real numbers. 3. Square $a + bi$ and set it equal to $4 + 3i$: $$ (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi = 4 + 3i $$ 4. Equate the real and imaginary parts: Real part: $a^2 - b^2 = 4$ Imaginary part: $2ab = 3$ 5. From the imaginary part, solve for $b$: $$ b = \frac{3}{2a} $$ 6. Substitute $b$ into the real part equation: $$ a^2 - \left( \frac{3}{2a} \right)^2 = 4 $$ $$ a^2 - \frac{9}{4a^2} = 4 $$ 7. Multiply both sides by $4a^2$ to clear the denominator: $$ 4a^4 - 9 = 16a^2 $$ 8. Rearranged: $$ 4a^4 - 16a^2 - 9 = 0 $$ 9. Let $x = a^2$, rewrite as: $$ 4x^2 - 16x - 9 = 0 $$ 10. Solve this quadratic equation: $$ x = \frac{16 \pm \sqrt{16^2 - 4 \cdot 4 \cdot (-9)}}{2 \cdot 4} = \frac{16 \pm \sqrt{256 + 144}}{8} = \frac{16 \pm \sqrt{400}}{8} = \frac{16 \pm 20}{8} $$ 11. Thus, two solutions for $x$: $$ x_1 = \frac{16 + 20}{8} = \frac{36}{8} = 4.5 $$ $$ x_2 = \frac{16 - 20}{8} = \frac{-4}{8} = -0.5 \text{ (discard negative since } a^2 \geq 0) $$ 12. So, $a^2 = 4.5$, hence $a = \pm \sqrt{4.5} = \pm \frac{3}{\sqrt{2}}$. 13. Find $b$ using $b = \frac{3}{2a}$: If $a = \frac{3}{\sqrt{2}}$, $$ b = \frac{3}{2 \cdot \frac{3}{\sqrt{2}}} = \frac{3}{\frac{6}{\sqrt{2}}} = \frac{3 \sqrt{2}}{6} = \frac{\sqrt{2}}{2} $$ If $a = -\frac{3}{\sqrt{2}}$, $$ b = \frac{3}{2 \cdot -\frac{3}{\sqrt{2}}} = -\frac{\sqrt{2}}{2} $$ 14. Therefore, the square roots are: $$ \sqrt{4+3i} = \pm \left( \frac{3}{\sqrt{2}} + \frac{\sqrt{2}}{2} i \right) $$ 15. Simplify $a$ by rationalizing denominator: $$ \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} $$ Final answer: $$ \boxed{ \pm \left( \frac{3\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \right) } $$