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

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

Evaluate

$$-\sqrt{3}+i\approx -1.732050808+i$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

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

Calculate 1+i to the power of 2 and get 2i.

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

Expand

$$-\sqrt{3}+i$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

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

Calculate 1+i to the power of 2 and get 2i.

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