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lyrics to here i am lord hymn - IX Labs

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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If \( a = -1 \), \( b = 1 \), \( (-1 + i)^2 = 1 - 2i + i^2 = -2i \) ✅
So the square roots are \( 1 - i \) and \( -1 + i \)
Now,
\[ z = \frac{-(1 + i) \pm (1 - i)}{2} \]
First solution:
\[ z = \frac{-1 - i + 1 - i}{2} = \frac{-2i}{2} = -i \]
Second solution:
\[ z = \frac{-1 - i -1 + i}{2} = \frac{-2}{2} = -1 \]
Thus, the solutions are \( z = -i \) and \( z = -1 \)
\(\boxed{z = -1 \text{ and } z = -i}\)

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