Question

(-1/2 + √3i/2)^2

please solve the problem
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Answer 1

Answer:

The answer is $-\frac{1}{2}-\frac{\sqrt 3}{2}i$

Step-by-step explanation:

$(-\frac{1}{2}+\frac{\sqrt 3i}{2})^2=(\frac{-1+\sqrt 3i}{2})^2\\=\frac{(-1+\sqrt 3i)^2}{4}=\frac{1+3i^2-2\sqrt 3i}{4}\\=\frac{1+3(-1)-2\sqrt 3}{4}=\frac{-2-2\sqrt 3i}{4}\\=\frac{-1-\sqrt 3i}{2}=-\frac{1}{2}-\frac{\sqrt 3}{2}i$