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

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

Evaluate

$$\sqrt{3}\left(-\frac{1}{4}+\frac{1}{4}i\right)+\left(\frac{1}{4}+\frac{1}{4}i\right)\approx -0.183012702+0.683012702i$$

Solution Steps

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

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

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}.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

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

Add 1 and 3 to get 4.

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

Divide \left(1+i\right)\left(1+i\sqrt{3}\right) by 4 to get \left(\frac{1}{4}+\frac{1}{4}i\right)\left(1+i\sqrt{3}\right).

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

Use the distributive property to multiply \frac{1}{4}+\frac{1}{4}i by 1+i\sqrt{3}.

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

Real Part

$$\frac{1 - \sqrt{3}}{4} = -0.1830127018922193$$