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

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

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}.

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

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.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -i and 3 to get -3i.

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

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

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

Subtract 2i from -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

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