\begin{align*} (1 + 3i)x + (2 - i)y & = 4 + 2i \\ x + 3x i + 2y - yi & = 4 + 2i \\ (x + 2y) + (3x - y)i & = 4 + 2i \\ \\ \text{Comparing } & \text{real parts,} \\ x + 2y & = 4 \phantom{00} \text{--- (1)} \\ \\ \text{Comparing } & \text{imaginary parts,} \\ 3x - y & = 2 \phantom{00} \text{--- (2)} \\ \\ \text{Solve (1) } & \text{& (2) by GC,} \\ x = {8 \over 7}, & \phantom{.} y = {10 \over 7} \end{align*}

\begin{align*} \text{Let } \pm \sqrt{3 + 4i} & = x + yi, \text{ where } x, y \in \mathbb{R} \\ \\ 3 + 4i & = (x + yi)^2 \\ 3 + 4i & = x^2 + 2(x)(yi) + (yi)^2 \\ 3 + 4i & = x^2 + 2xyi + y^2 i^2 \\ 3 + 4i & = x^2 + 2xyi + y^2 (-1) \\ 3 + 4i & = (x^2 - y^2) + 2xy i \\ \\ \text{Comparing } & \text{real parts,} \\ 3 & = x^2 - y^2 \phantom{00} \text{-- (1)} \\ \\ \text{Comparing } & \text{imaginary parts,} \\ 4 & = 2xy \\ 2 & = xy \\ {2 \over x} & = y \phantom{00} \text{-- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3 & = x^2 - \left(2 \over x\right)^2 \\ 3 & = x^2 - {4 \over x^2} \\ 3x^2 & = x^4 - 4 \\ 0 & = x^4 - 3x^2 - 4 \\ 0 & = (x^2)^2 - 3(x^2) - 4 \\ 0 & = (x^2 - 4)(x^2 + 1) \end{align*} \begin{align*} x^2 - 4 & = 0 &&& x^2 + 1 & = 0 \\ x^2 & = 4 &&& x^2 & = -1 \text{ (Reject, since } x \in \mathbb{R}) \\ x & = \pm 2 \end{align*} \begin{align*} \text{Substitute } & x = \pm 2 \text{ into (2),} \\ y & = {2 \over \pm 2} \\ y & = \pm 1 \\ \\ \therefore \sqrt{3 + 4i} & = 2 + i \\ -\sqrt{3 + 4i} & = -2 - i \end{align*}

\begin{align*} {6 + 10i \over 2i} & = {6 \over 2i} + {10i \over 2i} \\ & = {3 \over i} \times {i \over i} + 5 \\ & = {3i \over -1} + 5 \\ & = -3i + 5 \end{align*}

\begin{align*} {5 + 10i \over 4 + 3i} & = {5 + 10i \over 4 + 3i} \times {4 - 3i \over 4 - 3i} \phantom{000000} [\text{Conjugate of } 4 + 3i] \\ & = {(5 + 10i)(4 - 3i) \over (4 + 3i)(4 - 3i)} \\ & = {20 - 15i + 40i - 30i^2 \over (4)^2 - (3i)^2} \\ & = {20 + 25i + 30 \over 16 - (-9)} \\ & = {50 + 25i \over 25} \\ & = {50 \over 25} + {25i \over 25} \\ & = 2 + i \end{align*}