Solve frac{1+sqrt{3}i}{1-sqrt{3}i}

$$\frac { 1 + \sqrt { 3 } i } { 1 - \sqrt { 3 } i }$$

Evaluate

$$\frac{-1+\sqrt{3}i}{2}\approx -0.5+0.866025404i$$

Solution Steps

Rationalize the denominator of \frac{1+i\sqrt{3}}{1-i\sqrt{3}} by multiplying numerator and denominator by 1+i\sqrt{3}.

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

Consider \left(1-i\sqrt{3}\right)\left(1+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}.

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

Multiply 1+i\sqrt{3} and 1+i\sqrt{3} to get \left(1+i\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+i\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

$$\frac{-2+2i\sqrt{3}}{1+3}$$

Add 1 and 3 to get 4.

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

Real Part

$$-\frac{1}{2} = -0.5$$