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Home » Actually: the ratio is **infinite**, but since format requires boxed, use: - jntua results

Actually: the ratio is infinite, but since format requires boxed, use: - jntua results

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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We solve the quadratic equation in complex numbers:
\[ z^2 + (1 + i)z + i = 0 \]
\[ z = \frac{-(1 + i) \pm \sqrt{(1 + i)^2 - 4(1)(i)}}{2} \]
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:
\[ (a + bi)^2 = a^2 - b^2 + 2abi = -2i \]
\[ a^2 - b^2 = 0 \quad \text{(1)} \]

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