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First, compute the discriminant:
\[ (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]
\[ 4 \cdot i = 4i \]
\[ \text{Discriminant} = 2i - 4i = -2i \]
Now, compute \( \sqrt{-2i} \). Let \( \sqrt{-2i} = a + bi \), then:
\[ (a + bi)^2 = a^2 - b^2 + 2abi = -2i \]
\[ a^2 - b^2 = 0 \quad \text{(1)} \]
\[ 2ab = -2 \quad \text{(2)} \]
From (1): \( a^2 = b^2 \Rightarrow a = b \) or \( a = -b \)
Try \( a = b \):