Solve (1/2-frac{sqrt{3}i}{2})^2

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

Evaluate

$$\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

Solution Steps

Divide i\sqrt{3} by 2 to get \frac{1}{2}i\sqrt{3}.

$$\left(\frac{1}{2}-\frac{1}{2}i\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}-\frac{1}{2}i\sqrt{3}\right)^{2}.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{1}{4}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{1}{4}\times 3$$

Multiply -\frac{1}{4} and 3 to get -\frac{3}{4}.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{3}{4}$$

Subtract \frac{3}{4} from \frac{1}{4} to get -\frac{1}{2}.

$$-\frac{1}{2}-\frac{1}{2}i\sqrt{3}$$

Expand

$$\frac{-\sqrt{3}i-1}{2}$$

Solution Steps

Divide i\sqrt{3} by 2 to get \frac{1}{2}i\sqrt{3}.

$$\left(\frac{1}{2}-\frac{1}{2}i\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}-\frac{1}{2}i\sqrt{3}\right)^{2}.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{1}{4}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{1}{4}\times 3$$

Multiply -\frac{1}{4} and 3 to get -\frac{3}{4}.

$$\frac{1}{4}-\frac{1}{2}i\sqrt{3}-\frac{3}{4}$$

Subtract \frac{3}{4} from \frac{1}{4} to get -\frac{1}{2}.

$$-\frac{1}{2}-\frac{1}{2}i\sqrt{3}$$