Solve frac{2sqrt{2}+2sqrt{6}i}{sqrt{2}i+sqrt{6}}

$$\frac { 2 \sqrt { 2 } + 2 \sqrt { 6 } i } { \sqrt { 2 } i + \sqrt { 6 } }$$

Evaluate

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

Solution Steps

Rationalize the denominator of \frac{2\sqrt{2}+2i\sqrt{6}}{i\sqrt{2}+\sqrt{6}} by multiplying numerator and denominator by i\sqrt{2}-\sqrt{6}.

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{\left(i\sqrt{2}+\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}$$

Consider \left(i\sqrt{2}+\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{\left(i\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{i^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{-\left(\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{-2-\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{-2-6}$$

Subtract 6 from -2 to get -8.

$$\frac{\left(2\sqrt{2}+2i\sqrt{6}\right)\left(i\sqrt{2}-\sqrt{6}\right)}{-8}$$

Apply the distributive property by multiplying each term of 2\sqrt{2}+2i\sqrt{6} by each term of i\sqrt{2}-\sqrt{6}.

$$\frac{2i\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{6}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

The square of \sqrt{2} is 2.

$$\frac{2i\times 2-2\sqrt{2}\sqrt{6}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Multiply 2i and 2 to get 4i.

$$\frac{4i-2\sqrt{2}\sqrt{6}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{4i-2\sqrt{2}\sqrt{2}\sqrt{3}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{4i-2\times 2\sqrt{3}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Multiply -2 and 2 to get -4.

$$\frac{4i-4\sqrt{3}-2\sqrt{2}\sqrt{6}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{4i-4\sqrt{3}-2\sqrt{2}\sqrt{2}\sqrt{3}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{4i-4\sqrt{3}-2\times 2\sqrt{3}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Multiply -2 and 2 to get -4.

$$\frac{4i-4\sqrt{3}-4\sqrt{3}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

Combine -4\sqrt{3} and -4\sqrt{3} to get -8\sqrt{3}.

$$\frac{4i-8\sqrt{3}-2i\left(\sqrt{6}\right)^{2}}{-8}$$

The square of \sqrt{6} is 6.

$$\frac{4i-8\sqrt{3}-2i\times 6}{-8}$$

Multiply -2i and 6 to get -12i.

$$\frac{4i-8\sqrt{3}-12i}{-8}$$

Subtract 12i from 4i to get -8i.

$$\frac{-8i-8\sqrt{3}}{-8}$$

Multiply both numerator and denominator by -1.

$$\frac{8i+8\sqrt{3}}{8}$$

Real Part

$$\sqrt{3} = 1.732050808$$