Solve the equation $z^2 + 4z + 13 = 0$.
A$z = -2 \pm 6i$
B$z = -2 \pm 3i$CORRECT
C$z = -4 \pm 3i$
D$z = 2 \pm 3i$
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Why the answer is B, and why the others tempt you.
Using the quadratic formula, $z = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = \dfrac{-4 \pm \sqrt{-36}}{2} = \dfrac{-4 \pm 6i}{2} = -2 \pm 3i$. Forgetting to divide the imaginary part by 2 gives $-2 \pm 6i$.
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