Solve frac{5+2isqrt{3}}{3+isqrt{3}}

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

Evaluate

$$\frac{\sqrt{3}i}{12}+\frac{7}{4}\approx 1.75+0.144337567i$$

Solution Steps

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

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

Consider \left(3+i\sqrt{3}\right)\left(3-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}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{9-\left(i\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{9-i^{2}\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{9-\left(-\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{9-\left(-3\right)}$$

Multiply -1 and -3 to get 3.

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{9+3}$$

Add 9 and 3 to get 12.

$$\frac{\left(5+2i\sqrt{3}\right)\left(3-i\sqrt{3}\right)}{12}$$

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

$$\frac{15-5i\sqrt{3}+6i\sqrt{3}+2\left(\sqrt{3}\right)^{2}}{12}$$

Combine -5i\sqrt{3} and 6i\sqrt{3} to get i\sqrt{3}.

$$\frac{15+i\sqrt{3}+2\left(\sqrt{3}\right)^{2}}{12}$$

The square of \sqrt{3} is 3.

$$\frac{15+i\sqrt{3}+2\times 3}{12}$$

Multiply 2 and 3 to get 6.

$$\frac{15+i\sqrt{3}+6}{12}$$

Add 15 and 6 to get 21.

$$\frac{21+i\sqrt{3}}{12}$$

Real Part

$$\frac{7}{4} = 1\frac{3}{4} = 1.75$$