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

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

Evaluate

$$\frac{\sqrt{3}+5i}{4}\approx 0.433012702+1.25i$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square -i.

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

Subtract -1 from 3 to get 4.

$$\frac{\left(2+i\sqrt{3}\right)\left(\sqrt{3}+i\right)}{4}$$

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

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

The square of \sqrt{3} is 3.

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

Add 2i and 3i to get 5i.

$$\frac{2\sqrt{3}+5i-\sqrt{3}}{4}$$

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

$$\frac{\sqrt{3}+5i}{4}$$

Real Part

$$\frac{\sqrt{3}}{4} = 0.4330127018922193$$