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

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

Evaluate

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

Solution Steps

Use the distributive property to multiply \sqrt{3}+i by 2-2i.

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

Use the distributive property to multiply \sqrt{3}-i by 2+2i.

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

Rationalize the denominator of \frac{\left(2-2i\right)\sqrt{3}+\left(2+2i\right)}{\left(2+2i\right)\sqrt{3}+\left(2-2i\right)} by multiplying numerator and denominator by \left(2+2i\right)\sqrt{3}+\left(-2+2i\right).

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

Consider \left(\left(2+2i\right)\sqrt{3}+\left(2-2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(\left(2+2i\right)\sqrt{3}\right)^{2}.

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

Calculate 2+2i to the power of 2 and get 8i.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{8i\left(\sqrt{3}\right)^{2}-\left(2-2i\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{8i\times 3-\left(2-2i\right)^{2}}$$

Multiply 8i and 3 to get 24i.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{24i-\left(2-2i\right)^{2}}$$

Calculate 2-2i to the power of 2 and get -8i.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{24i-\left(-8i\right)}$$

Multiply -1 and -8i to get 8i.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{24i+8i}$$

Add 24i and 8i to get 32i.

$$\frac{\left(\left(2-2i\right)\sqrt{3}+\left(2+2i\right)\right)\left(\left(2+2i\right)\sqrt{3}+\left(-2+2i\right)\right)}{32i}$$

Apply the distributive property by multiplying each term of \left(2-2i\right)\sqrt{3}+\left(2+2i\right) by each term of \left(2+2i\right)\sqrt{3}+\left(-2+2i\right).

$$\frac{8\left(\sqrt{3}\right)^{2}+8i\sqrt{3}+8i\sqrt{3}-8}{32i}$$

The square of \sqrt{3} is 3.

$$\frac{8\times 3+8i\sqrt{3}+8i\sqrt{3}-8}{32i}$$

Multiply 8 and 3 to get 24.

$$\frac{24+8i\sqrt{3}+8i\sqrt{3}-8}{32i}$$

Combine 8i\sqrt{3} and 8i\sqrt{3} to get 16i\sqrt{3}.

$$\frac{24+16i\sqrt{3}-8}{32i}$$

Subtract 8 from 24 to get 16.

$$\frac{16+16i\sqrt{3}}{32i}$$

Real Part

$$\frac{\sqrt{3}}{2} = 0.8660254037844386$$