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}.

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

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

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

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

Add 1 and 3 to get 4.

$$\frac{\left(-1-i\sqrt{3}\right)\left(-1-i\sqrt{3}\right)}{4}$$

Use the distributive property to multiply -1-i\sqrt{3} by -1-i\sqrt{3}.

$$\frac{-\left(-1-i\sqrt{3}\right)-i\left(-1-i\sqrt{3}\right)\sqrt{3}}{4}$$

To find the opposite of -1-i\sqrt{3}, find the opposite of each term.

$$\frac{-\left(-1\right)+i\sqrt{3}-i\left(-1-i\sqrt{3}\right)\sqrt{3}}{4}$$

The opposite of -1 is 1.

$$\frac{1+i\sqrt{3}-i\left(-1-i\sqrt{3}\right)\sqrt{3}}{4}$$

Use the distributive property to multiply -i by -1-i\sqrt{3}.

$$\frac{1+i\sqrt{3}+\left(i-\sqrt{3}\right)\sqrt{3}}{4}$$

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

$$\frac{1+i\sqrt{3}+i\sqrt{3}-\left(\sqrt{3}\right)^{2}}{4}$$

The square of \sqrt{3} is 3.

$$\frac{1+i\sqrt{3}+i\sqrt{3}-3}{4}$$

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

$$\frac{1+2i\sqrt{3}-3}{4}$$

Subtract 3 from 1 to get -2.

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

Real Part

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