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

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

Evaluate

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

Solution Steps

To raise \frac{1+i\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

$$\frac{-2+2i\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

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

Expand

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

Solution Steps

To raise \frac{1+i\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

$$\frac{-2+2i\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

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