Solve frac{2+6sqrt{3}i}{5+sqrt{3}i}

$$\frac { 2 + 6 \sqrt { 3 } i } { 5 + \sqrt { 3 } i }$$

Evaluate

$$1+\sqrt{3}i\approx 1+1.732050808i$$

Solution Steps

Rationalize the denominator of \frac{2+6i\sqrt{3}}{5+i\sqrt{3}} by multiplying numerator and denominator by 5-i\sqrt{3}.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{\left(5+i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}$$

Consider \left(5+i\sqrt{3}\right)\left(5-i\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{5^{2}-\left(i\sqrt{3}\right)^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{25-\left(i\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{25-i^{2}\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{25-\left(-\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{25-\left(-3\right)}$$

Multiply -1 and -3 to get 3.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{25+3}$$

Add 25 and 3 to get 28.

$$\frac{\left(2+6i\sqrt{3}\right)\left(5-i\sqrt{3}\right)}{28}$$

Apply the distributive property by multiplying each term of 2+6i\sqrt{3} by each term of 5-i\sqrt{3}.

$$\frac{10-2i\sqrt{3}+30i\sqrt{3}+6\left(\sqrt{3}\right)^{2}}{28}$$

Combine -2i\sqrt{3} and 30i\sqrt{3} to get 28i\sqrt{3}.

$$\frac{10+28i\sqrt{3}+6\left(\sqrt{3}\right)^{2}}{28}$$

The square of \sqrt{3} is 3.

$$\frac{10+28i\sqrt{3}+6\times 3}{28}$$

Multiply 6 and 3 to get 18.

$$\frac{10+28i\sqrt{3}+18}{28}$$

Add 10 and 18 to get 28.

$$\frac{28+28i\sqrt{3}}{28}$$

Real Part

$$1$$