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Solution:
Let (\sqrt{3 - 4i} = a + bi) with (a, b \in \mathbb{R}).
Squaring both sides: ((a + bi)^2 = a^2 - b^2 + 2abi = 3 - 4i).
Equating real and imaginary parts:
[
\begin{cases}
a^2 - b^2 = 3
2ab = -4 \implies ab = -2
\end{cases}
]
From (ab = -2), we have (b = -2/a) (for (a \neq 0)). Substituting into the first equation:
[
a^2 - \frac{4}{a^2} = 3 \implies a^4 - 3a^2 - 4 = 0
]
Let (x = a^2), giving (x^2 - 3x - 4 = 0), so ((x - 4)(x + 1) = 0). Since (a) is real, (a^2 = 4 \implies a = \pm 2).
If (a = 2), then (b = -1).
If (a = -2), then (b = 1).
Thus, the square roots are (2 - i) and (-2 + i).
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